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NCERT Exemplar · Class 9 Science Sound

20 questions · 20 still being checked

Long Answer Questions 18–20 (part 3 of 3)

  1. Exercise 18

    Represent graphically by two separate diagrams in each case
    (i)
    Two sound waves having the same amplitude but different frequencies?
    (ii)
    Two sound waves having the same frequency but different amplitudes.
    (iii)
    Two sound waves having different amplitudes and also different
    wavelengths.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Amplitude fixes a curve's crest height; frequency (or wavelength, on a distance axis) fixes its crest spacing. Each pair is drawn as two curves in its own band. NCERT_Solution_Class9_Science_Exemplar_Ch12_Q18 Answer: (i) equal height, unequal spacing — different frequency. (ii) equal spacing, unequal height — different amplitude. (iii) both height and spacing unequal — different amplitude and wavelength.

  2. Exercise 19

    Establish the relationship between speed of sound, its wavelength and
    frequency. If velocity of sound in air is 340\displaystyle 340 m s−1\displaystyle s^{-1}, calculate
    (i)
    wavelength when frequency is 256\displaystyle 256 Hz.
    (ii)
    frequency when wavelength is 0.85\displaystyle 0.85 m.

    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
    Derivation of formula v = ν λ. (i) $\displaystyle 340$ = $\displaystyle 256$ λ λ = $\displaystyle 1.33$ m. (ii) $\displaystyle 340$ = ν ($\displaystyle 0.85$) ν = $\displaystyle 400$ Hz
    In one time period \(\displaystyle T\) a wave advances one wavelength \(\displaystyle \lambda\), so
    \[v = \frac{\text{distance}}{\text{time}} = \frac{\lambda}{T} \]
    Since \(\displaystyle f = \dfrac{1}{T}\),
    \[v = f\lambda \]
    Given \(\displaystyle v = 340\ \text{m s}^{-1}\).
    (i)
    \(\displaystyle f = 256\ \text{Hz}\):
    \[\lambda = \frac{v}{f} = \frac{340\ \text{m s}^{-1}}{256\ \text{Hz}} = 1.33\ \text{m} \]
    (ii)
    \(\displaystyle \lambda = 0.85\ \text{m}\):
    \[f = \frac{v}{\lambda} = \frac{340\ \text{m s}^{-1}}{0.85\ \text{m}} = 400\ \text{Hz} \]
    Answer: \(\displaystyle v = f\lambda\); (i) \(\displaystyle \lambda = 1.33\ \text{m}\) (ii) \(\displaystyle f = 400\ \text{Hz}\).

  3. Exercise 20

    Draw a curve showing density or pressure variations with respect to distance for a disturbance produced by sound. Mark the position of compression and rarefaction on this curve. Also define wavelengths and time period using this curve.

    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
    Wavelength is the distance between two consecutive compressions or two consecutive rarefactions. Time period is the time taken to travel the distance between any two consecutive compressions or rarefactions from a fixed point.
    NCERT_Solution_Class9_Science_Exemplar_Ch12_Q20 The crests are compressions (maximum density/pressure) and the troughs are rarefactions (minimum density/pressure), marked on the curve above.Wavelength \(\displaystyle \lambda\): the distance between two consecutive compressions (or two consecutive rarefactions), marked above between the two crests.Time period \(\displaystyle T\): the time for one complete compression–rarefaction cycle to pass a fixed point — the time for the disturbance to travel one wavelength, \[v = \frac{\lambda}{T} \;\Rightarrow\; T = \frac{\lambda}{v}, \] where \(\displaystyle v\) is the speed of sound.Answer: \(\displaystyle \lambda\) = distance between two consecutive compressions/rarefactions; \(\displaystyle T = \lambda/v\) is the time for one such cycle to cross a fixed point.