SolveItClass 9 · NCERT

NCERT Solutions · Class 9 Science Sound Waves: Characteristics and Applications

28 questions · 16 still being checked

Revise, Reflect, Refine 10.2–10.3 (part 3 of 21)

  1. Exercise 10.2

    For a sound wave propagating in a medium, increasing its frequency will increase its (i) wavelength (ii) speed (iii) number of compressions per second (iv) time period

    Not cross-checked

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

    NCERT’s answer
    (iii)
    (iii) number of compressions per second.
    NCERT_Solution_Class9_Science_Ch10_RRR_Q10-2
    Frequency is the number of density oscillations passing a point each second, so raising it means more compressions arrive every second.
    (ii) is wrong: in a given medium the speed depends on the medium (and its temperature and humidity), not on the source or its frequency.
    (i) is wrong: since \(\displaystyle v = \nu \times \lambda \) stays fixed, a larger \(\displaystyle \nu \) forces a smaller \(\displaystyle \lambda \).
    (iv) is wrong: \(\displaystyle T = \dfrac{1}{\nu} \), so a higher frequency gives a shorter time period.

  2. Exercise 10.3

    If 20\displaystyle 20 compressions pass a point in 4\displaystyle 4 seconds, the frequency is (i) 80\displaystyle 80 Hz (ii) 5\displaystyle 5 Hz (iii) 10\displaystyle 10 Hz (iv) 0.2\displaystyle 0.2 Hz
    NCERT’s answer
    $\displaystyle 5$ Hz
    (ii) $\displaystyle 5$ Hz.
    Each compression passing the point counts as one complete density oscillation there.
    \(\displaystyle \nu = \dfrac{\text{number of oscillations}}{\text{time taken}} = \dfrac{20}{4\ \text{s}} = 5\ \text{Hz} \).
    $\displaystyle 5$ Hz is below $\displaystyle 20$ Hz, so this would in fact be an infrasonic wave — inaudible to humans.