NCERT Solutions for Class 9th Science Chapter 10 .6.1 Wavelength, frequency and time period — Pause and Ponder

Book page 19510 Updated on2026-09-08

Q1.

Conduct Activity 10.1 once again with a thick rubber band and then with a thin rubber band. Does the thin rubber band vibrate faster than the thick rubber band? If yes, how do the frequency and time period of the sound produced by the thin rubber band differ from that of the thick rubber band?

Answer

Yes — for the same length and the same tension, the thin band vibrates faster than the thick one.

Band	Rate of vibration	Frequency ν	Time period T = 1/ν	Sound heard
Thin band	Faster	Higher	Shorter	Shriller — high pitch
Thick band	Slower	Lower	Longer	Deeper — low pitch

ν = 1/T (Eq. 10.1)
If the thin band completes 400 oscillations in 1 s, ν = 400 Hz and T = 1/400 s = 0.0025 s
If the thick band completes 200 oscillations in 1 s, ν = 200 Hz and T = 1/200 s = 0.005 s

Why thickness matters: a thicker band carries more mass on every centimetre of its length. The same restoring force must now accelerate more mass, so the band is slower to swing back and forth — each complete oscillation takes longer. A longer time period means a lower frequency, because ν and T are inverses of each other.

Did you know? This is why a sitar or veena carries strings of different thicknesses side by side: the thick strings give the low notes and the thin ones the high notes, all at roughly the same tension.

Q2.

If the frequency of a sound wave produced by an oscillating piston of a long tube filled with air is 20 Hz, then how many oscillations does the piston complete per minute?

Answer

1200 oscillations per minute.

Frequency ν = number of oscillations ÷ time taken
∴ number of oscillations = ν × time
ν = 20 Hz = 20 s⁻¹, time = 1 minute = 60 s
number of oscillations = 20 s⁻¹ × 60 s = 1200

Why the units work out: hertz means 'per second', so 20 Hz literally reads '20 oscillations every second'. Multiplying by 60 s cancels the seconds and leaves a pure number — a count of oscillations, which is exactly what was asked for. Always convert the minute to seconds before multiplying; that single step is where most mistakes happen.

Tip: 20 Hz sits right at the lower edge of the human audible range (20 Hz – 20 kHz). Anything slower than this piston would produce infrasound, which we cannot hear at all.

Q3.

For the sound wave represented by the graph shown in Fig. 10.19, what is half of its wavelength?

Answer

Half the wavelength is 1.5 cm (= 0.015 m).

Read the graph first. The curve starts on the average-density line at 0, rises to a crest, comes back down through the line, dips to a trough, and returns to the line at the mark labelled 3.0 cm. That completes one full density oscillation — so one whole wavelength is measured on the printed scale as

λ = distance between two consecutive crests (or two consecutive troughs)
From Fig. 10.19: λ = 3.0 cm
∴ λ/2 = 3.0 cm ÷ 2 = 1.5 cm = 0.015 m

Reading Fig. 10.19: one complete density oscillation spans 0 to 3.0 cm, so λ = 3.0 cm and half the wavelength is 1.5 cm.

Why half a wavelength is worth naming: it is exactly the distance from the centre of a compression to the centre of the next rarefaction. Half a wavelength away from a crest, the density has swung from its maximum right down to its minimum.

Tip: a common slip is to measure from a crest to the next trough and call that the wavelength. That gap is only λ/2. Always measure crest-to-crest or trough-to-trough.

Was this helpful? Report an error