SolveItNCERT · CBSE Boards

NCERT Exemplar · Class 9 Science Sound

20 questions · 20 still being checked

Short Answer Questions 10–17 (part 2 of 3)

  1. Exercise 10

    The given graph (Fig.12.2) shows the displacement versus time relation for a disturbance travelling with velocity of 1500\displaystyle 1500 m s−1\displaystyle s^{-1}. Calculate the wavelength of the disturbance. NCERT_Question_Class9_Science_Exemplar_Ch12_Q10

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    From the graph (Fig. $\displaystyle 12.2$), successive crests occur at \(\displaystyle t = 1\), \(\displaystyle 3\), \(\displaystyle 5\ \mu\text{s}\), so the time period is \[T = 2\ \mu\text{s} = 2\times10^{-6}\ \text{s} \] Wavelength: \[\lambda = vT = 1500\ \text{m s}^{-1} \times 2\times10^{-6}\ \text{s} = 3\times10^{-3}\ \text{m} \]Answer: \(\displaystyle \lambda = 3\times10^{-3}\ \text{m} = 3\ \text{mm}\)NCERT prints: \(\displaystyle \lambda = v/\nu = 5\times10^{5}\ \text{m}\) — it repeats the frequency's value instead of dividing by it; \(\displaystyle v/\nu = 1500\ \text{m s}^{-1} / (5\times10^{5}\ \text{Hz}) = 3\times10^{-3}\ \text{m}\).

  2. Exercise 11

    Which of the above two graphs (a) and (b) (Fig.12.3) representing the human voice is likely to be the male voice? Give reason for your answer. NCERT_Question_Class9_Science_Exemplar_Ch12_Q11

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Graph (a) represents the male voice. Usually the male voice has less pitch (or frequency) as compared to female.
    Graph (a) completes fewer oscillations than graph (b) over the same time interval, so \[f_a < f_b. \] Frequency and pitch move together, \(\displaystyle f\downarrow \Rightarrow \text{pitch}\downarrow\), and a male voice is deeper (lower-pitched) than a female voice.Answer: Graph (a) is the male voice — it has the lower frequency of the two curves.

  3. Exercise 12

    A girl is sitting in the middle of a park of dimension 12\displaystyle 12 m × 12\displaystyle 12 m. On the left side of it there is a building adjoining the park and on right side of the park, there is a road adjoining the park. A sound is produced on the road by a cracker. Is it possible for the girl to hear the echo of this sound? Explain your answer.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    If the time gap between the original sound and reflected sound received by the listener is around $\displaystyle 0.1$ s, only then the echo can be heard. The minimum distance travelled by the reflected sound wave for the distinctly listening the echo = velocity of sound × time interval ; $\displaystyle 344$ × $\displaystyle 0.1$ ; $\displaystyle 34.4$ m But in this case the distance travelled by the sound reflected from the building and then reaching to the girl will be ($\displaystyle 6$ + $\displaystyle 6$) = $\displaystyle 12$ m, which is much smaller than the required distance. Therefore, no echo can be heard.
    A distinct echo needs the reflected sound to lag the direct one by \(\displaystyle \Delta t\ge0.1\text{ s}\). The girl, at the park's centre, is \(\displaystyle 6\text{ m}\) from the cracker on the road; the reflected sound travels cracker \(\displaystyle \to\) building wall \(\displaystyle \to\) girl, a path \(\displaystyle 12\text{ m}+6\text{ m}=18\text{ m}\) long. \[t_{direct}=\frac{6\text{ m}}{344\text{ m s}^{-1}}=0.017\text{ s} \] \[t_{echo}=\frac{18\text{ m}}{344\text{ m s}^{-1}}=0.052\text{ s} \] \[\Delta t=0.052\text{ s}-0.017\text{ s}=0.035\text{ s} \] Answer: No — \(\displaystyle \Delta t=0.035\text{ s}<0.1\text{ s}\), too short for the ear to hear the reflection as a separate echo.

  4. Exercise 13

    Why do we hear the sound produced by the humming bees while the sound of vibrations of pendulum is not heard?

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Humming bees produce sound by vibrating their wings which is in the audible range. In case of pendulum the frequency is below $\displaystyle 20$ Hz which does not come in the audible range.
    Human hearing spans \(\displaystyle 20\) to \(\displaystyle 20{,}000\ \text{Hz}\); a source outside this band goes unheard. \[20\ \text{Hz} \le f_{\text{bee wings}} \le 20{,}000\ \text{Hz} \;\Rightarrow\; \text{audible} \] \[f_{\text{pendulum}} < 20\ \text{Hz} \;\Rightarrow\; \text{infrasonic, inaudible} \] A bee's wingbeat lies in the audible band, so it is heard; a pendulum's swing is far slower, so it is not.Answer: The bee's wing-beat frequency lies within \(\displaystyle 20\)-\(\displaystyle 20{,}000\ \text{Hz}\) (audible); the pendulum's is far below \(\displaystyle 20\ \text{Hz}\) (infrasonic), so only the bee is heard.

  5. Exercise 14

    If any explosion takes place at the bottom of a lake, what type of shock waves in water will take place?

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Longitudinal waves.
    Water is a fluid: it cannot sustain a shearing (sideways) stress, so it transmits only longitudinal -- compression and rarefaction -- disturbances, never transverse ones.Answer: Longitudinal (compressional) shock waves, spreading outward through the water in all directions from the blast.

  6. Exercise 15

    Sound produced by a thunderstorm is heard 10\displaystyle 10 s after the lightning is seen. Calculate the approximate distance of the thunder cloud. (Given speed of sound = 340\displaystyle 340 m s−1\displaystyle s^{-1}.)

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    s ; $\displaystyle 340$ m \(\displaystyle s^{-1}\) × $\displaystyle 10$ s = $\displaystyle 3400$ m or $\displaystyle 3.4$ km.
    Light from the flash reaches the observer in a negligible time; only the sound's travel time counts. \[d=v\,t \] \[d=340\text{ m s}^{-1}\times10\text{ s} \] \[d=3400\text{ m}=3.4\text{ km} \] Answer: the thundercloud is about \(\displaystyle 3.4\text{ km}\) away.

  7. Exercise 16

    For hearing the loudest ticking sound heard by the ear, find the angle x in the Fig.12.4. NCERT_Question_Class9_Science_Exemplar_Ch12_Q16

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    ∠i = ∠r, so x = $\displaystyle 90$° - ∠r = $\displaystyle 90$°- $\displaystyle 50$° = $\displaystyle 40$°
    Sound obeys the same law of reflection as light: the angle of incidence equals the angle of reflection, both measured from the Normal. In Fig. $\displaystyle 12.4$ the \(\displaystyle 50^\circ\) arc runs from the wall to the incident tube, not from the Normal, so \[\angle i = 90^\circ - 50^\circ = 40^\circ \] \[x = \angle r = \angle i = 40^\circ \] Answer: \(\displaystyle x = 40^\circ\).

  8. Exercise 17

    Why is the ceiling and wall behind the stage of good conference halls or concert halls made curved?

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Ceiling and walls are made curved so that sound after reflection reaches the target audience.
    A curved (concave) wall obeys \(\displaystyle \theta_i=\theta_r\); with the source S at its focus, every reflected ray leaves parallel to the hall's axis, spreading the sound evenly over the whole audience. A flat wall instead reflects a diverging beam that fades with distance, leaving the back rows faint.Answer: Curving the wall, source at its focus, turns the reflection parallel — every seat gets near-equal loudness.