SolveItClass 9 · NCERT

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

28 questions · 16 still being checked

Pause and Ponder 10.8–10.9 (part 12 of 21)

  1. Exercise 10.8

    If the frequency of a sound wave produced by an oscillating piston of a long tube filled with air is 20\displaystyle 20 Hz, then how many oscillations does the piston complete per minute?
    NCERT’s answer
    $\displaystyle 1200$ oscillations
    $\displaystyle 1200$ oscillations per minute.
    Frequency is the number of oscillations per unit time, so \(\displaystyle \nu = 20\ \text{Hz} \) means the piston completes $\displaystyle 20$ oscillations every second.
    One minute is $\displaystyle 60$ s, so \(\displaystyle \text{oscillations} = 20\ \text{s}^{-1} \times 60\ \text{s} = 1200 \).
    Each of those $\displaystyle 1200$ oscillations sends out one compression and one rarefaction into the tube.

  2. Exercise 10.9

    For the sound wave represented by the graph shown in Fig. 10.19\displaystyle 10.19, what is half of its wavelength? The amplitude The amount of sound energy passing through a unit area perpendicularNCERT_Question_Class9_Science_Ch10_PP_Q10-9
    NCERT’s answer
    1.$\displaystyle 5$ cm
    Half the wavelength is $\displaystyle 1.5$ cm, so the full wavelength is \(\displaystyle 3.0\ \text{cm} \).
    On the distance axis of Fig. $\displaystyle 10.19$ the marks \(\displaystyle 0,\ 1.5,\ 3.0,\ 4.5\ \text{cm} \) fall alternately on a crest and a trough.
    Wavelength is the distance between two consecutive crests (or two consecutive troughs): \(\displaystyle \lambda = 3.0\ \text{cm} - 0 = 3.0\ \text{cm} \).
    \(\displaystyle \dfrac{\lambda}{2} = \dfrac{3.0\ \text{cm}}{2} = 1.5\ \text{cm} \) — which is exactly the crest-to-next-trough spacing you can read straight off the graph.