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NCERT Solutions · Class 11 Physics Oscillations

18 questions · 6 still being checked

Exercises 13.1–13.10 (part 1 of 2)

  1. Exercise 13.1

    Which of the following examples represent periodic motion?
    (a)
    A swimmer completing one (return) trip from one bank of a river to the other and back.
    (b)
    A freely suspended bar magnet displaced from its N-S direction and released.
    (c)
    A hydrogen molecule rotating about its centre of mass.
    (d)
    An arrow released from a bow.

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    NCERT’s answer
    (b), (c)
    Periodic motion means the motion repeats itself after equal intervals of time — not just "back and forth," but back and forth (or around and around) again and again, with the same time interval every time. Check each case against that definition rather than against how the motion looks.(a) The swimmer's round trip. The swimmer goes from one bank to the other and back — once. As described, this is a single to-and-fro journey, not a motion that keeps repeating at equal intervals. Nothing says the swimmer repeats the trip, and even if the swimmer did it again, there is no reason the return trip takes the same time as the first (currents, fatigue, strokes all vary). With no fixed interval of repetition, this motion is not periodic.(b) The freely suspended bar magnet. A magnet hanging freely aligns with the north-south direction because of the restoring torque exerted on it (the same idea behind a compass needle). Displace it and let go: the magnetic restoring torque pulls it back, it overshoots, the torque reverses, and it swings back — the same swing, over and over, taking the same time each cycle. This is oscillation about the N-S line, and it recurs at equal time intervals, so it is periodic.(c) The rotating hydrogen molecule. A rotating rigid body sweeps through the same set of orientations again after each full turn. If the molecule spins about its centre of mass at a steady angular velocity \(\displaystyle \omega \), it returns to the same orientation after every rotation, and each rotation takes the same time \(\displaystyle T = \dfrac{2\pi}{\omega} \). That equal-time repetition is exactly what periodic means — so this is periodic, even though it is not a to-and-fro motion like (b). (Periodic motion need not be oscillatory: oscillation requires moving back and forth about a mean position, but a steady rotation is periodic without ever reversing direction.)(d) The arrow released from a bow. Once released, the arrow flies off in one direction and does not return to repeat that flight. There is exactly one pass, no recurrence, so this is not periodic.The distinguishing test throughout is repetition at equal time intervals — a single occurrence, however similar it looks to a repeating one, fails that test, while both to-and-fro oscillation and steady rotation pass it.Answer: (b) and (c) represent periodic motion; (a) and (d) do not.

  2. Exercise 13.2

    Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? the rotation of earth about its axis. motion of an oscillating mercury column in a U-tube. motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point. general vibrations of a polyatomic molecule about its equilibrium position.

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    NCERT’s answer
    (b)
    and (c): SHM; (a) and (d) represent periodic but not SHM [A polyatomic molecule has a number of natural frequencies; so in general, its vibration is a superposition of SHM’s of a number of different frequencies. This superposition is periodic but not SHM].
    A motion is periodic if it repeats after a fixed time \(\displaystyle T\); it is simple harmonic only if, in addition, the restoring force pulling it back to one fixed equilibrium point is directly proportional to the displacement from that point and opposite in direction, \(\displaystyle F = -kx\). Every SHM is periodic, but not every periodic motion is SHM — you have to check the force law in each case, not just whether the motion repeats.(a) Rotation of the Earth about its axis — periodic, not SHM. The Earth returns to the same orientation every $\displaystyle 24$ hours, so the motion is periodic. But SHM requires a to-and-fro motion about a fixed equilibrium position under a restoring force \(\displaystyle F=-kx\). Rotation is not oscillatory at all — the Earth keeps turning steadily in the same sense, there is no equilibrium point it is being pulled back toward, and no restoring force. So this is periodic but not SHM (in fact, not oscillatory in any sense).(b) The oscillating mercury column in a U-tube — SHM. Let the mercury (density \(\displaystyle \rho\), total column length \(\displaystyle l\), tube cross-section \(\displaystyle A\)) be in equilibrium with equal levels in both limbs. Push the level down by \(\displaystyle x\) in one limb; by conservation of volume it rises by \(\displaystyle x\) in the other, so the height difference between the two columns becomes \(\displaystyle 2x\).By the law of hydrostatic pressure, this height difference produces an unbalanced pressure \[\Delta p = \rho g (2x) \] where \(\displaystyle g\) is the acceleration due to gravity. This pressure acts over area \(\displaystyle A\), giving a net restoring force \[F = -\Delta p \, A = -2\rho g A\, x . \] The minus sign is because the force always pushes the mercury back toward the level position, opposite to the displacement \(\displaystyle x\). This has exactly the form \(\displaystyle F=-kx\) with \(\displaystyle k = 2\rho g A\), so (neglecting viscosity) the motion is SHM. Its angular frequency, using the total mercury mass \(\displaystyle m=\rho A l\), is \[\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{2\rho g A}{\rho A l}} = \sqrt{\dfrac{2g}{l}} . \](c) The ball bearing in a smooth curved bowl — SHM. Near its lowest point, a smooth curved bowl behaves like an arc of a circle of some radius \(\displaystyle R\), the same geometry as a simple pendulum bob swinging near the bottom of its arc. If the ball is displaced through a small arc length \(\displaystyle x\) from the lowest point (equivalent to a small angle \(\displaystyle \theta = x/R\) measured from the centre of curvature), gravity supplies a tangential restoring component \[F = -mg\sin\theta \approx -mg\left(\frac{x}{R}\right) = -\left(\frac{mg}{R}\right)x , \] using the small-angle approximation \(\displaystyle \sin\theta \approx \theta\) (valid because the ball is released only "slightly above" the lowest point, so \(\displaystyle x\) stays small). This is again of the form \(\displaystyle F=-kx\), with \(\displaystyle k = mg/R\), so the motion is nearly SHM, with \[\omega = \sqrt{\dfrac{g}{R}} . \] The word "nearly" matters here, just as with the mercury column: real bowls and real fluids depart from the idealized force law once the amplitude grows, so this is simple harmonic only in the small-displacement limit.(d) General vibrations of a polyatomic molecule about equilibrium — periodic, not SHM. A molecule with \(\displaystyle N\) atoms has several independent vibrational normal modes (each individually obeying \(\displaystyle F=-kx\) about the equilibrium bond lengths/angles, and so each individually SHM), but each normal mode has its own characteristic frequency. "General vibration" means several of these normal modes are excited at once, so the actual motion of an atom is a superposition of two or more SHMs of different frequencies. Such a superposition still repeats itself after a fixed time (it is periodic, with period equal to the least common multiple of the individual periods) — but the resultant waveform is no longer a single sine curve, so it is not itself simple harmonic. Only a diatomic molecule, which has just one vibrational mode, vibrates in pure SHM; a general polyatomic molecule does not.Answer: (b) the oscillating mercury column and (c) the ball bearing in the smooth curved bowl are (nearly) simple harmonic motion; (a) the Earth's rotation and (d) the general vibration of a polyatomic molecule are periodic but not simple harmonic.

  3. Exercise 13.3

    Fig. 13.18\displaystyle 13.18 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion) ? Fig. 18.18\displaystyle 18.18NCERT_Question_Class11_Physics_Ch13_Q13-3
    NCERT’s answer
    (b)
    and (d) are periodic, each with a period of $\displaystyle 2$ s; (a) and (c) are not periodic. [Note in (c), repetition of merely one position is not enough for motion to be periodic; the entire motion during one period must be repeated successively].
    The test for periodic motion is repetition of the whole graph, not just of a value: the motion is periodic only if there is a fixed time \(\displaystyle T\) such that \(\displaystyle x(t+T)=x(t)\) for every instant \(\displaystyle t\), and the period is the smallest such \(\displaystyle T\). Here \(\displaystyle x\) is the displacement read off the vertical axis and \(\displaystyle t\) is the time read off the horizontal axis (in seconds). A curve that merely comes back to \(\displaystyle x=0\) every so often is not enough — the shape of the journey in between must repeat too.Plot (a) — not periodic. Reading the curve: it starts at the origin and climbs continuously to the right, getting steeper as \(\displaystyle t\) grows, and it never turns back. Every value of \(\displaystyle x\) is reached once and only once, so no \(\displaystyle T\) can satisfy \(\displaystyle x(t+T)=x(t)\). This is a particle moving off in one direction (speeding up), not oscillating.Plot (b) — periodic, \(\displaystyle T = 2\ \mathrm{s}\). The curve is a saw-tooth: a nearly vertical fall from a maximum down to a minimum, then a slow climb (with a small kink partway up) back to the next maximum, and that unit is drawn again and again, identically — four falls are visible in the figure. To get the period, locate the falls against the printed marks on the \(\displaystyle t\)-axis: one fall sits between the marks \(\displaystyle -3\) and \(\displaystyle -1\), i.e. at \(\displaystyle t = -2\ \mathrm{s}\), and another sits between the marks \(\displaystyle 1\) and \(\displaystyle 3\), i.e. at \(\displaystyle t = +2\ \mathrm{s}\). Exactly one more fall (the one at the origin) lies between those two, so that stretch of the axis holds two complete saw-teeth: \[T = \frac{t_2 - t_1}{\text{number of repeats}} = \frac{(+2\ \mathrm{s}) - (-2\ \mathrm{s})}{2} = \frac{4\ \mathrm{s}}{2} = 2\ \mathrm{s}. \] So \(\displaystyle x(t+2\ \mathrm{s}) = x(t)\) for every \(\displaystyle t\): periodic with period \(\displaystyle 2\ \mathrm{s}\).Plot (c) — not periodic. This is the trap in the question. The curve does come down and touch \(\displaystyle x = 0\) at \(\displaystyle t = 1, 4, 7, 10, 13\ \mathrm{s}\) — at every one of the printed marks, equally spaced \(\displaystyle 3\ \mathrm{s}\) apart — which tempts you to write \(\displaystyle T = 3\ \mathrm{s}\). But look at what the particle does between those touches; each $\displaystyle 3$-second stretch is a different curve:
    \(\displaystyle 1\to 4\ \mathrm{s}\): a small ledge, then one rounded hump peaking near \(\displaystyle t \approx 3\ \mathrm{s}\).
    \(\displaystyle 4\to 7\ \mathrm{s}\): only low ripples, less than half the height of that hump.
    \(\displaystyle 7\to 10\ \mathrm{s}\): a tall sharp spike peaking near \(\displaystyle t \approx 8.4\ \mathrm{s}\) — the highest point of the whole graph, roughly \(\displaystyle 1.5\) times the hump near \(\displaystyle 3\ \mathrm{s}\) — followed by a small bump.
    \(\displaystyle 10\to 13\ \mathrm{s}\): a single broad hump of middling height, peaking near \(\displaystyle t \approx 12\ \mathrm{s}\).
    Since the peaks have four different heights and four different shapes, \(\displaystyle x(t+3\ \mathrm{s}) \neq x(t)\), and no other interval works either, because no two stretches of this graph can be laid on top of each other. Repeating the return to zero is not repeating the motion: non-periodic.Plot (d) — periodic, \(\displaystyle T = 2\ \mathrm{s}\). This is a smooth sine-shaped curve. Read off the crossings: it cuts the \(\displaystyle t\)-axis at every marked second, \(\displaystyle t = -3, -2, -1, 0, 1, 2, 3\ \mathrm{s}\), with a crest halfway between one crossing and the next and a trough halfway between the following pair. One complete pattern is crest plus trough, so take two successive crossings in the same direction: the curve crosses zero going upward at \(\displaystyle t = 0\) (crest at \(\displaystyle 0.5\ \mathrm{s}\), trough at \(\displaystyle 1.5\ \mathrm{s}\)) and next crosses zero going upward at \(\displaystyle t = 2\ \mathrm{s}\). Hence \[T = (2\ \mathrm{s}) - (0\ \mathrm{s}) = 2\ \mathrm{s}. \] Note that a single crossing-to-crossing gap of \(\displaystyle 1\ \mathrm{s}\) is only half a cycle — the displacement there has flipped sign, so \(\displaystyle x(t+1\ \mathrm{s}) = -x(t)\), not \(\displaystyle x(t)\).Answer: Plots (b) and (d) represent periodic motion, each with period \(\displaystyle T = 2\ \mathrm{s}\); plots (a) and (c) are non-periodic — (a) rises steadily and never repeats, and (c) returns to \(\displaystyle x = 0\) every \(\displaystyle 3\ \mathrm{s}\) but traces a different shape between every pair of returns.

  4. Exercise 13.4

    Which of the following functions of time represent
    (a)
    simple harmonic,
    (b)
    periodic but not simple harmonic, and
    (c)
    non-periodic motion? Give period for each case of periodic motion (ω is any positive constant): (a) sin ωt - cos ωt (b) sin3\displaystyle sin^{3} ωt (c) 3\displaystyle 3 cos (π/4\displaystyle 4 - 2\displaystyle 2ωt)
    (d)
    cos ωt + cos 3\displaystyle 3ωt + cos 5\displaystyle 5ωt
    (e)
    exp (-ω2\displaystyle ω^{2}t2\displaystyle t^{2})
    (f)
    1\displaystyle 1 + ωt + ω2\displaystyle ω^{2}t2\displaystyle t^{2}
    NCERT’s answer
    (a)
    Simple harmonic, T = ($\displaystyle 2$π/ω); (b) periodic, T =($\displaystyle 2$π/ω) but not simple harmonic; (c) simple harmonic, T = (π/ω); (d) periodic, T = ($\displaystyle 2$π/ω) but not simple harmonic; (e) non-periodic; (f) non-periodic (physically not acceptable as the function → ∞ as t → ∞.
    A function of time is simple harmonic only when it collapses to a single sine or cosine term — one amplitude, one frequency, one phase. If a function repeats itself after a fixed time but needs more than one frequency to write it down (even if it started life as a single sine raised to a power), it is periodic but not SHM. If it never repeats at all, it is non-periodic.The formal test for SHM is Newton's second law applied to a restoring force: \(\displaystyle \dfrac{d^2x}{dt^2} = -\omega^2 x\), acceleration proportional to displacement and directed back toward zero. Any \(\displaystyle x(t)\) that can be written as \(\displaystyle A\sin(\omega t+\phi)\) (a single sinusoid) automatically satisfies this. A sum of sinusoids of different frequencies does not — each term separately satisfies its own such equation with its own \(\displaystyle \omega\), so the sum obeys no single equation of that form.(a) \(\displaystyle \sin\omega t - \cos\omega t\)Combine the two terms into one sinusoid using \(\displaystyle a\sin\theta + b\cos\theta = R\sin(\theta+\phi)\), with \(\displaystyle R=\sqrt{a^2+b^2}\). Here \(\displaystyle a=1,\ b=-1\), so \(\displaystyle R=\sqrt2\), and \[\sin\omega t - \cos\omega t = \sqrt2\sin\!\left(\omega t - \frac{\pi}{4}\right). \] Check: \(\displaystyle \sqrt2\sin(\omega t-\pi/4)=\sqrt2\left[\sin\omega t\cos\frac{\pi}{4}-\cos\omega t\sin\frac{\pi}{4}\right]=\sin\omega t-\cos\omega t\), which matches. This is a single sinusoid of amplitude \(\displaystyle \sqrt2\) and angular frequency \(\displaystyle \omega\) — simple harmonic motion, with period \[T=\frac{2\pi}{\omega}. \](b) \(\displaystyle \sin^3\omega t\)This looks like plain SHM because it is "just" a sine function, but raising a sine to a power creates new frequencies — that is the step most people miss. Use the triple-angle identity \(\displaystyle \sin3\theta = 3\sin\theta - 4\sin^3\theta\), rearranged: \[\sin^3\omega t = \frac{3\sin\omega t - \sin3\omega t}{4}. \] This is a superposition of two sinusoids at different frequencies, \(\displaystyle \omega\) and \(\displaystyle 3\omega\), so it cannot be reduced to a single \(\displaystyle A\sin(\omega t+\phi)\): it is periodic but not simple harmonic. Its two component periods are \(\displaystyle 2\pi/\omega\) and \(\displaystyle 2\pi/(3\omega)\); since the first is exactly three times the second, the whole expression repeats with the longer period, \[T=\frac{2\pi}{\omega}. \](c) \(\displaystyle 3\cos\!\left(\dfrac{\pi}{4}-2\omega t\right)\)Cosine is an even function, \(\displaystyle \cos(-x)=\cos x\), so \[3\cos\!\left(\frac{\pi}{4}-2\omega t\right) = 3\cos\!\left(2\omega t-\frac{\pi}{4}\right). \] This is a single sinusoid of amplitude $\displaystyle 3$ and angular frequency \(\displaystyle 2\omega\) — simple harmonic motion. Its period is set by the angular frequency actually multiplying \(\displaystyle t\), which is \(\displaystyle 2\omega\), not \(\displaystyle \omega\): \[T=\frac{2\pi}{2\omega}=\frac{\pi}{\omega}. \](d) \(\displaystyle \cos\omega t + \cos3\omega t + \cos5\omega t\)Three sinusoids at three different frequencies, \(\displaystyle \omega,\,3\omega,\,5\omega\), added together. No single frequency describes the sum, so it is periodic but not simple harmonic (this is the kind of anharmonic, "many-overtone" periodic wave you get from a plucked string or a non-sinusoidal oscillator). The individual periods are \(\displaystyle 2\pi/\omega,\ 2\pi/(3\omega),\ 2\pi/(5\omega)\); the first is $\displaystyle 3$ times the second and $\displaystyle 5$ times the third, so the whole sum first repeats after \[T=\frac{2\pi}{\omega}. \](e) \(\displaystyle \exp(-\omega^2t^2)\)This is a Gaussian (bell-shaped) curve centred at \(\displaystyle t=0\): it rises to a single maximum and falls monotonically to zero as \(\displaystyle t\to\pm\infty\), taking every value only once on each side of the peak. A periodic function must repeat, \(\displaystyle f(t+T)=f(t)\) for the same fixed \(\displaystyle T\) at every \(\displaystyle t\); this function never returns to a value it has already left, so no such \(\displaystyle T\) exists. This is non-periodic motion (it describes a pulse dying away, not an oscillation).(f) \(\displaystyle 1+\omega t+\omega^2t^2\)This is a quadratic (parabolic) function of time. As \(\displaystyle t\) grows, the \(\displaystyle \omega^2t^2\) term dominates and the value grows without bound in either direction — it is monotonic on each side of its vertex and never returns to a previous value on a repeating cycle. This is non-periodic motion.Answer: (a) SHM, \(\displaystyle T=2\pi/\omega\). (b) Periodic, not SHM, \(\displaystyle T=2\pi/\omega\). (c) SHM, \(\displaystyle T=\pi/\omega\). (d) Periodic, not SHM, \(\displaystyle T=2\pi/\omega\). (e) Non-periodic. (f) Non-periodic.

  5. Exercise 13.5

    A particle is in linear simple harmonic motion between two points, A and B, 10\displaystyle 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at the end A, at the end B, at the mid-point of AB going towards A, at 2\displaystyle 2 cm away from B going towards A, at 3\displaystyle 3 cm away from A going towards B, and at 4\displaystyle 4 cm away from B going towards A.
    NCERT’s answer
    (a)
    $\displaystyle 0$, +, + ; (b) $\displaystyle 0$, -, - ; (c) -, $\displaystyle 0,0$ ; (d) -, -, - ; (e) +, +, + ; (f ) -, -, -.
    Acceleration and force in SHM always point toward the mean position; velocity's sign only tracks which way the particle happens to be moving at that instant — these are two independent rules, so each point on the path needs its own check.Put an axis along AB with the midpoint O (the mean, equilibrium position) as the origin, and take the direction from A to B as positive \(\displaystyle x\). Since AB is $\displaystyle 10$ cm, the amplitude is \[a = \frac{AB}{2} = 5\ \text{cm}, \] so A sits at \(\displaystyle x = -5\) cm and B at \(\displaystyle x = +5\) cm.Newton's second law for SHM gives the restoring force as \(\displaystyle F = -kx\), so \[F = ma = -kx \implies a = -\omega^2 x,\qquad \omega^2 = \frac{k}{m}, \] where \(\displaystyle x\) is the signed displacement from the mean position O, \(\displaystyle m\) is the particle's mass, and \(\displaystyle \omega\) is the angular frequency. This says two things at once:
    Acceleration (and hence force, since \(\displaystyle F=ma\) with \(\displaystyle m>0\)) always points from wherever the particle is straight back toward O, with size proportional to how far it is from O. It never depends on which way the particle is moving — only on which side of O it sits.
    Velocity has no such rule. Its sign is simply positive if the particle is moving toward B, negative if it is moving toward A, regardless of position.
    Now check each point.At end A. Here \(\displaystyle x=-5\) cm, an extreme of the motion, so the particle is momentarily at rest: velocity \(\displaystyle =0\). It is on the negative side of O, so the restoring force pulls it toward O, i.e. toward B — the positive direction. Acceleration is positive, and so is the force.At end B. Here \(\displaystyle x=+5\) cm, the other extreme, so again velocity \(\displaystyle =0\). Being on the positive side of O, the restoring pull is toward O, i.e. toward A — the negative direction. Acceleration is negative, and so is the force.At the mid-point, moving toward A. This is O itself, \(\displaystyle x=0\), so \(\displaystyle a=-\omega^2(0)=0\) and hence \(\displaystyle F=0\) as well (this is where the particle has its maximum speed, even though the force on it is zero). The particle is moving toward A, the negative direction, so velocity is negative.At $\displaystyle 2$ cm from B, moving toward A. B is $\displaystyle 5$ cm from O, so a point $\displaystyle 2$ cm short of B is \(\displaystyle 5-2=3\) cm from O, on B's (positive) side: \(\displaystyle x=+3\) cm. Since \(\displaystyle a=-\omega^2(3)\) is negative, both acceleration and force point toward O, i.e. toward A — negative. The particle's motion is also toward A, so velocity is negative too.At $\displaystyle 3$ cm from A, moving toward B. A point $\displaystyle 3$ cm past A is \(\displaystyle 5-3=2\) cm from O, on A's (negative) side: \(\displaystyle x=-2\) cm. Then \(\displaystyle a=-\omega^2(-2)=+2\omega^2\) is positive, so acceleration and force point toward O, i.e. toward B — positive. The motion is also toward B, so velocity is positive.At $\displaystyle 4$ cm from B, moving toward A. A point $\displaystyle 4$ cm short of B is \(\displaystyle 5-4=1\) cm from O, on B's (positive) side: \(\displaystyle x=+1\) cm. Then \(\displaystyle a=-\omega^2(1)\) is negative, so acceleration and force point toward O, i.e. toward A — negative. The motion is toward A, so velocity is negative.A point people trip on: acceleration's sign is fixed the moment you know the particle's position, not its direction of travel — that's why a particle can be moving toward A while its acceleration also points toward A (as in the last two cases, because both those points lie on B's side of O), or moving toward B while its acceleration points toward B (the second-to-last case, because that point lies on A's side of O). Neither is a coincidence of direction matching; it's the geometry of which side of the mean position the particle is on.Answer: With A→B taken as positive — at A: velocity $\displaystyle 0$, acceleration positive, force positive. At B: velocity $\displaystyle 0$, acceleration negative, force negative. At the mid-point moving toward A: velocity negative, acceleration $\displaystyle 0$, force 0. At $\displaystyle 2$ cm from B moving toward A: velocity negative, acceleration negative, force negative. At $\displaystyle 3$ cm from A moving toward B: velocity positive, acceleration positive, force positive. At $\displaystyle 4$ cm from B moving toward A: velocity negative, acceleration negative, force negative.

  6. Exercise 13.6

    Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion? a = 0.7x a = -200x2\displaystyle 200x^{2} a = -10x a = 100x3\displaystyle 100x^{3}

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    NCERT’s answer
    (c)
    represents a simple harmonic motion.
    The signature of SHM is a straight-line law: acceleration must be a negative constant times displacement, and nothing else — no square, no cube, and not a positive constant either.For simple harmonic motion, Newton's second law combined with the restoring-force condition gives the defining relation \[a = -\omega^{2}x \] where \(\displaystyle \omega\) is the angular frequency (a positive constant) and \(\displaystyle x\) is the displacement from the equilibrium position. Two things must both hold:
    \(\displaystyle a\) must be directly proportional to \(\displaystyle x\) (only the first power of \(\displaystyle x\), nothing higher),
    the constant of proportionality must be negative, because the force (and hence the acceleration) always has to point back toward the equilibrium position, opposite to the displacement. A positive sign would mean the acceleration pushes the particle further away, which is unstable motion, not oscillation.
    Now check each relation against \(\displaystyle a = -\omega^{2}x\):(a) \(\displaystyle a = 0.7x\) This is linear in \(\displaystyle x\), but the coefficient is \(\displaystyle +0.7\), not negative. The acceleration points the same way as the displacement, so the particle is driven away from equilibrium rather than pulled back. This cannot be SHM.(b) \(\displaystyle a = -200x^{2}\) The sign is right, but the acceleration depends on \(\displaystyle x^{2}\), not \(\displaystyle x\). Squaring destroys the proportionality to \(\displaystyle x\) itself (also, \(\displaystyle x^2\) is never negative, so this acceleration would always point the same way regardless of which side of equilibrium the particle is on — it cannot represent a symmetric restoring force). This cannot be SHM.(c) \(\displaystyle a = -10x\) This is linear in \(\displaystyle x\) with a negative constant of proportionality. Matching it to \(\displaystyle a = -\omega^{2}x\) gives \[\omega^{2} = 10 \ \text{s}^{-2} \] which is a perfectly good positive constant, so this satisfies the SHM condition exactly.(d) \(\displaystyle a = 100x^{3}\) Here the acceleration depends on \(\displaystyle x^{3}\), a higher power of \(\displaystyle x\), not a direct proportionality. This cannot be SHM.Only relation (c) has the form \(\displaystyle a = -(\text{positive constant})x\), so only (c) describes simple harmonic motion, with \(\displaystyle \omega^{2} = 10\ \text{s}^{-2}\).Answer: Only \(\displaystyle a = -10x\) represents simple harmonic motion (matching \(\displaystyle a=-\omega^2 x\) with \(\displaystyle \omega^2 = 10\ \text{s}^{-2}\)); the other three do not, because \(\displaystyle 0.7x\) has the wrong sign and \(\displaystyle -200x^{2}\), \(\displaystyle 100x^{3}\) are not linear in \(\displaystyle x\).

  7. Exercise 13.7

    The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (ωt + φ ). If the initial (t = 0\displaystyle 0) position of the particle is 1\displaystyle 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle ? The angular frequency of the particle is π s−1\displaystyle s^{-1}. If instead of the cosine function, we choose the sine function to describe the SHM : x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.

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    NCERT’s answer
    A = $\displaystyle 2$ cm, φ = $\displaystyle 7$π/$\displaystyle 4$; B = $\displaystyle 2$ cm, a = π/4.
    Match the position and the velocity at \(\displaystyle t=0\) to two equations, then combine them like the two legs of a right triangle — the amplitude is the hypotenuse.The particle's motion is \(\displaystyle x(t) = A\cos(\omega t + \varphi)\). Differentiating gives the velocity, \[v(t) = \frac{dx}{dt} = -A\omega\sin(\omega t + \varphi). \]Put in the two initial conditions, \(\displaystyle x(0) = 1\ \text{cm}\) and \(\displaystyle v(0) = \omega\ \text{cm/s}\): \[A\cos\varphi = 1\ \text{cm} \qquad \text{(i)} \] \[-A\omega\sin\varphi = \omega\ \text{cm/s} \qquad \text{(ii)} \]In (ii) the angular frequency \(\displaystyle \omega\) appears on both sides and cancels — this is the step people rush past: the numerical value \(\displaystyle \omega = \pi\ \text{s}^{-1}\) given in the problem never actually enters the amplitude or the phase, because \(\displaystyle v(0)\) was chosen proportional to \(\displaystyle \omega\) on purpose. Dividing (ii) through by \(\displaystyle \omega\), \[A\sin\varphi = -1\ \text{cm}. \qquad \text{(ii}'\text{)} \]Square (i) and (ii\(\displaystyle '\)) and add, using \(\displaystyle \cos^2\varphi + \sin^2\varphi = 1\): \[A^2\cos^2\varphi + A^2\sin^2\varphi = (1\ \text{cm})^2 + (-1\ \text{cm})^2 \] \[A^2 = 2\ \text{cm}^2 \implies A = \sqrt{2}\ \text{cm} \approx 1.4\ \text{cm}. \]For the phase, divide (ii\(\displaystyle '\)) by (i): \[\tan\varphi = \frac{A\sin\varphi}{A\cos\varphi} = \frac{-1}{1} = -1. \] Here is the second trap: \(\displaystyle \tan\varphi = -1\) alone only fixes \(\displaystyle \varphi\) up to a multiple of \(\displaystyle \pi\) — you need the signs of \(\displaystyle \cos\varphi\) and \(\displaystyle \sin\varphi\) separately to pick the right quadrant. From (i), \(\displaystyle \cos\varphi = 1/\sqrt2 > 0\); from (ii\(\displaystyle '\)), \(\displaystyle \sin\varphi = -1/\sqrt2 < 0\). Positive cosine and negative sine place \(\displaystyle \varphi\) in the fourth quadrant, so \[\varphi = -\frac{\pi}{4}\ \text{rad}\ \ (-45^\circ), \] not \(\displaystyle +3\pi/4\), which is the other solution of \(\displaystyle \tan\varphi=-1\) but has both cosine and sine negative.Now repeat the same steps for the sine description, \(\displaystyle x(t) = B\sin(\omega t + \alpha)\), whose velocity is \[v(t) = B\omega\cos(\omega t + \alpha). \]The same initial conditions \(\displaystyle x(0)=1\ \text{cm}\), \(\displaystyle v(0) = \omega\ \text{cm/s}\) give \[B\sin\alpha = 1\ \text{cm} \qquad \text{(iii)} \] \[B\cos\alpha = 1\ \text{cm} \qquad \text{(iv)} \] (the \(\displaystyle \omega\) cancels again, the same way as before). Squaring and adding, \[B^2 = (1\ \text{cm})^2 + (1\ \text{cm})^2 = 2\ \text{cm}^2 \implies B = \sqrt{2}\ \text{cm} \approx 1.4\ \text{cm}, \] the same amplitude as before — it has to be, since both expressions describe the identical physical oscillation, just written with a different reference angle.For the phase, dividing (iii) by (iv): \[\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{1}{1} = 1, \] and here both \(\displaystyle \sin\alpha = 1/\sqrt2\) and \(\displaystyle \cos\alpha = 1/\sqrt2\) are positive, placing \(\displaystyle \alpha\) in the first quadrant: \[\alpha = \frac{\pi}{4}\ \text{rad}\ \ (45^\circ). \]As a check, using \(\displaystyle \sin\left(\theta + \dfrac{\pi}{2}\right) = \cos\theta\): \(\displaystyle \cos\left(\omega t - \dfrac{\pi}{4}\right) = \sin\left(\omega t - \dfrac{\pi}{4} + \dfrac{\pi}{2}\right) = \sin\left(\omega t + \dfrac{\pi}{4}\right)\) — the cosine form with \(\displaystyle \varphi=-\pi/4\) and the sine form with \(\displaystyle \alpha = \pi/4\) are the same function, confirming both results are consistent with each other.Answer: With \(\displaystyle x = A\cos(\omega t+\varphi)\): \(\displaystyle A = \sqrt{2}\ \text{cm} \approx 1.4\ \text{cm}\), \(\displaystyle \varphi = -\pi/4\ \text{rad} = -45^\circ\). With \(\displaystyle x = B\sin(\omega t+\alpha)\): \(\displaystyle B = \sqrt{2}\ \text{cm} \approx 1.4\ \text{cm}\), \(\displaystyle \alpha = \pi/4\ \text{rad} = 45^\circ\).

  8. Exercise 13.8

    A spring balance has a scale that reads from 0\displaystyle 0 to 50\displaystyle 50 kg. The length of the scale is 20\displaystyle 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6\displaystyle 0.6 s. What is the weight of the body ?
    NCERT’s answer
    $\displaystyle 219$ N
    The spring balance's own full-scale reading tells you its spring constant — the oscillation period then tells you the actual mass, which need not match the static reading.A spring balance is just a spring with a mass-calibrated scale: a hanging weight stretches the spring by an amount \(\displaystyle x \), and Hooke's law, \(\displaystyle F = kx \), converts that stretch into a force (which the dial then relabels as a mass reading, using \(\displaystyle g \)). Here \(\displaystyle k \) is the spring constant and \(\displaystyle x \) is the extension from the unloaded position.Step $\displaystyle 1$ — Find the spring constant from the scale's range.The scale runs from $\displaystyle 0$ to $\displaystyle 50$ kg over a length of $\displaystyle 20$ cm, so the full $\displaystyle 20$ cm of extension corresponds to the weight of a $\displaystyle 50$ kg mass:\[x_{\max} = 20\ \text{cm} = 0.20\ \text{m}, \qquad F_{\max} = (50\ \text{kg})(9.8\ \text{m/s}^2) = 490\ \text{N} \]\[k = \frac{F_{\max}}{x_{\max}} = \frac{490\ \text{N}}{0.20\ \text{m}} = 2450\ \text{N/m} \]Step $\displaystyle 2$ — Use the oscillation period to find the body's actual mass.When the loaded pan is displaced and released, it executes simple harmonic motion because the restoring force is \(\displaystyle F = -kx \) (Hooke's law again — the same law that calibrated the scale). The period of a mass-spring system is\[T = 2\pi\sqrt{\frac{m}{k}} \]where \(\displaystyle m \) is the mass on the pan and \(\displaystyle T = 0.6\ \text{s} \) is the given period. Squaring and solving for \(\displaystyle m \):\[m = \frac{kT^2}{4\pi^2} = \frac{(2450\ \text{N/m})(0.6\ \text{s})^2}{4\pi^2} = \frac{(2450)(0.36)}{39.48}\ \text{kg} = \frac{882}{39.48}\ \text{kg} \]\[m \approx 22.3\ \text{kg} \]This is the one step where it is easy to go wrong: the period of oscillation depends on the body's actual inertial mass, not on whatever number the dial happens to be pointing at while it swings — the dial reading is only meaningful once the system is back at rest.Step $\displaystyle 3$ — Convert mass to weight.Weight is a force, not a mass, so it must be reported in newtons, using \(\displaystyle W = mg \):\[W = mg = (22.34\ \text{kg})(9.8\ \text{m/s}^2) = 218.9\ \text{N} \]The weight acts vertically downward, as it always does for a hanging body.Significant figures. The given data ($\displaystyle 50$ kg, $\displaystyle 20$ cm, $\displaystyle 0.6$ s) carry at most two or three significant figures, so the result should not be reported with more precision than that. Rounding once, at the end:\[W \approx 219\ \text{N} \]Answer: W ≈ $\displaystyle 219$ N (directed vertically downward), corresponding to a mass of about $\displaystyle 22.3$ kg

  9. Exercise 13.9

    A spring having with a spring constant 1200\displaystyle 1200 N m−1\displaystyle m^{-1} is mounted on a horizontal table as shown in Fig. 13.19. A mass of 3\displaystyle 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0\displaystyle 2.0 cm and released. Fig. 13.19\displaystyle 13.19 Determine
    (i)
    the frequency of oscillations,
    (ii)
    maximum acceleration of the mass, and
    (iii)
    the maximum speed of the mass.
    NCERT’s answer
    Frequency $\displaystyle 3.2$ \(\displaystyle s^{-1}\); maximum acceleration of the mass $\displaystyle 8.0$ m \(\displaystyle s^{-2}\); maximum speed of the mass $\displaystyle 0.4$ m \(\displaystyle s^{-1}\).
    A mass on a spring executes simple harmonic motion, and every quantity you need — frequency, acceleration, speed — follows from just two numbers: the angular frequency \(\displaystyle \omega \) and the amplitude \(\displaystyle a \).The restoring force from the spring is \(\displaystyle F = -kx \) (Hooke's law), and Newton's second law \(\displaystyle F = ma \) then gives the equation of motion for SHM, whose angular frequency is \[\omega = \sqrt{\dfrac{k}{m}} \] where \(\displaystyle k \) is the spring constant and \(\displaystyle m \) is the mass attached to the spring — not its weight, its mass.Here \(\displaystyle k = 1200 \ \text{N m}^{-1} \) and \(\displaystyle m = 3 \ \text{kg} \), so \[\omega = \sqrt{\dfrac{1200}{3}} = \sqrt{400} = 20 \ \text{rad s}^{-1} \](i) Frequency of oscillationThe frequency \(\displaystyle \nu \) (oscillations per second) is related to the angular frequency by \(\displaystyle \omega = 2\pi\nu \), so \[\nu = \dfrac{\omega}{2\pi} = \dfrac{20}{2\pi} = \dfrac{20}{6.283} = 3.18 \ \text{Hz} \] Rounding to three significant figures (matching the three-figure precision of \(\displaystyle k = 1200 \)), \(\displaystyle \nu \approx 3.18 \ \text{Hz} \).(ii) Maximum accelerationDisplacing the mass by $\displaystyle 2.0$ cm and releasing it makes that displacement the amplitude of the oscillation: \(\displaystyle a = 2.0 \ \text{cm} = 2.0 \times 10^{-2} \ \text{m} \).In SHM the acceleration is \(\displaystyle A(t) = -\omega^2 x(t) \), which is largest in magnitude when the displacement is largest — that is, at the extreme position \(\displaystyle x = \pm a \), not while passing through the centre. So \[A_{\max} = \omega^2 a = (20)^2 \times (2.0 \times 10^{-2}) = 400 \times 0.02 = 8.0 \ \text{m s}^{-2} \] This is where the mass is momentarily at rest (about to reverse direction), which is easy to mix up with the point of maximum speed — those two happen at opposite ends of the motion.(iii) Maximum speedThe velocity in SHM is \(\displaystyle v(t) = -\omega a\sin(\omega t) \), which is largest in magnitude when \(\displaystyle \sin(\omega t) = \pm 1 \), i.e. as the mass crosses the centre (equilibrium position) — the opposite point from where the acceleration peaks. So \[v_{\max} = \omega a = 20 \times (2.0 \times 10^{-2}) = 0.40 \ \text{m s}^{-1} \]Answer: frequency \(\displaystyle \approx 3.18 \ \text{Hz} \); maximum acceleration \(\displaystyle = 8.0 \ \text{m s}^{-2} \) (at the extreme displacement); maximum speed \(\displaystyle = 0.40 \ \text{m s}^{-1} \) (at the equilibrium position).

  10. Exercise 13.10

    In Exercise 13.9\displaystyle 13.9, let us take the position of mass when the spring is unstreched as x = 0\displaystyle 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0\displaystyle 0), the mass is
    (a)
    at the mean position,
    (b)
    at the maximum stretched position, and
    (c)
    at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
    NCERT’s answer
    (a)
    x = $\displaystyle 2$ sin 20t (b) x = $\displaystyle 2$ cos 20t (c) x = - $\displaystyle 2$ cos 20t where x is in cm. These functions differ neither in amplitude nor frequency. They differ in initial phase.
    Only the initial phase changes here — the angular frequency and the amplitude are fixed by the spring and the mass, not by when you start the stopwatch.For a block on a spring obeying Hooke's law, Newton's second law gives \[m\frac{d^2x}{dt^2} = -kx \quad\Rightarrow\quad \frac{d^2x}{dt^2} = -\left(\frac{k}{m}\right)x, \] which is the SHM equation \(\displaystyle \dfrac{d^2x}{dt^2} = -\omega^2 x \) with angular frequency \[\omega = \sqrt{\frac{k}{m}}. \]From Exercise $\displaystyle 13.9$, the spring constant is \(\displaystyle k = 1200\ \text{N m}^{-1} \) and the mass is \(\displaystyle m = 3\ \text{kg} \), so \[\omega = \sqrt{\frac{1200\ \text{N m}^{-1}}{3\ \text{kg}}} = \sqrt{400\ \text{s}^{-2}} = 20\ \text{rad s}^{-1}. \]The mass was pulled out by \(\displaystyle 2.0\ \text{cm} \) and released from rest, so that is the amplitude of every oscillation that follows: \(\displaystyle A = 2.0\ \text{cm} \), no matter which instant we choose to call \(\displaystyle t = 0 \).The general solution of the SHM equation is \[x(t) = A\sin(\omega t + \phi), \] where \(\displaystyle \phi \) is the initial phase — fixed only by the position (and direction of motion) at \(\displaystyle t = 0 \). This is the one place the three cases below differ.(a) Mass at the mean position at \(\displaystyle t = 0 \). Here \(\displaystyle x(0) = 0 \), so \(\displaystyle A\sin\phi = 0 \), giving \(\displaystyle \phi = 0 \) or \(\displaystyle \pi \). Taking the mass to be moving toward the stretched side (the positive x-direction, by the sign convention given) as it crosses the mean position picks out \(\displaystyle \phi = 0 \): \[x(t) = (2.0\ \text{cm})\,\sin(20\,t), \] with \(\displaystyle t \) in seconds, and the argument \(\displaystyle 20t \) evaluated in radians, not degrees — an easy slip on a calculator.(b) Mass at the maximum stretched position at \(\displaystyle t = 0 \). Here \(\displaystyle x(0) = +A \), and the mass is momentarily at rest there, so \(\displaystyle \dot{x}(0) = 0 \). A cosine satisfies both conditions directly: \[x(t) = (2.0\ \text{cm})\,\cos(20\,t). \] Written in the sine form above, \(\displaystyle \cos(20t) = \sin(20t + \pi/2) \), so \(\displaystyle \phi = \pi/2 \).(c) Mass at the maximum compressed position at \(\displaystyle t = 0 \). Now \(\displaystyle x(0) = -A \), again with \(\displaystyle \dot{x}(0) = 0 \): \[x(t) = -(2.0\ \text{cm})\,\cos(20\,t) = (2.0\ \text{cm})\,\sin(20\,t - \pi/2), \] so \(\displaystyle \phi = -\pi/2 \) (equivalently \(\displaystyle 3\pi/2 \)).Comparing the three. The angular frequency \(\displaystyle \omega = 20\ \text{rad s}^{-1} \) (equivalently \(\displaystyle \nu = \omega/2\pi \approx 3.2\ \text{Hz} \), period \(\displaystyle T = 2\pi/\omega \approx 0.31\ \text{s} \)) and the amplitude \(\displaystyle A = 2.0\ \text{cm} \) come only from \(\displaystyle k \), \(\displaystyle m \), and how far the spring was stretched before release — none of that depends on the arbitrary instant we label \(\displaystyle t = 0 \), so all three functions share them exactly. What changes from one case to the next is only the initial phase \(\displaystyle \phi \): \(\displaystyle 0 \), \(\displaystyle \pi/2 \), and \(\displaystyle -\pi/2 \) rad respectively — a shift along the time axis, not any change in the size or shape of the oscillation.Answer: \(\displaystyle x_a(t) = 2.0\sin(20t)\ \text{cm} \), \(\displaystyle x_b(t) = 2.0\cos(20t)\ \text{cm} \), \(\displaystyle x_c(t) = -2.0\cos(20t)\ \text{cm} \) (t in s); all three share \(\displaystyle \omega = 20\ \text{rad s}^{-1} \) and \(\displaystyle A = 2.0\ \text{cm} \) and differ only in initial phase (\(\displaystyle 0 \), \(\displaystyle \pi/2 \), \(\displaystyle -\pi/2 \) rad).