Waves — Practice

Question 1

A wave is described by y(x,t) = 0.02 sin(4x - 2t) m. Find its wavelength and period.

Accepted Answer

k = 4 per m, so lambda = 2 pi/4 = 1.57 m. omega = 2 per s, so T = 2 pi/2 = 3.14 s.

Question 2

A string of linear mass density 0.05 kg/m is under a tension of 20 N. Find the speed of a transverse wave on it.

Accepted Answer

v = sqrt(T/mu) = sqrt(20/0.05) = sqrt(400) = 20 m/s.

Question 3

Two sound waves of frequency 256 Hz and 260 Hz are superposed. Find the beat frequency.

Accepted Answer

nu_beat = |260 - 256| = 4 Hz.

Question 4

Find the speed of sound in a gas at pressure 1.0 x 10^5 Pa and density 1.3 kg/m^3, given gamma = 1.4.

Accepted Answer

v = sqrt(gamma P/rho) = sqrt(1.4 x 10^5/1.3) = sqrt(107692) = 328 m/s.

Question 5

A wave is given by y(x,t) = 5 sin(3x + 6t) cm. Find its speed and direction of propagation.

Accepted Answer

The +x and +t terms share a plus sign, so the wave travels in the -x direction. v = omega/k = 6/3 = 2 cm/s.

Question 6

A stationary wave on a string has nodes 15 cm apart. Find the wavelength.

Accepted Answer

Node spacing = lambda/2 = 15 cm, so lambda = 30 cm.

Question 7

A string of length 1.2 m, fixed at both ends, carries transverse waves at speed 60 m/s. Find its fundamental frequency.

Accepted Answer

nu_1 = v/(2L) = 60/(2 x 1.2) = 60/2.4 = 25 Hz.

Question 8

A pipe closed at one end has length 0.5 m. Taking the speed of sound as 340 m/s, find its fundamental frequency.

Accepted Answer

nu_1 = v/(4L) = 340/(4 x 0.5) = 340/2 = 170 Hz.

Question 9

A resonance tube (closed at the water-level end) resonates with a 512 Hz tuning fork at air-column lengths of 16.4 cm and 50.5 cm (successive resonances). Find the speed of sound in air, and estimate the end correction.

Accepted Answer

Successive resonance lengths differ by lambda/2: 50.5 - 16.4 = 34.1 cm, so lambda = 68.2 cm = 0.682 m. v = lambda x nu = 0.682 x 512 = 349 m/s. End correction e = lambda/4 - L1 = 17.05 - 16.4 = 0.65 cm.

Question 10

Two tuning forks A and B produce 4 beats per second. When B is loaded with a bit of wax (which lowers its frequency), the beat frequency decreases to 2 beats per second. If the frequency of A is 256 Hz, find the original frequency of B.

Accepted Answer

Since loading B lowers its frequency and the beat frequency DECREASED, B's frequency was already below A's (moving further below would have increased the gap). So nu_A - nu_B = 4, giving nu_B = 256 - 4 = 252 Hz.

Practice — Waves

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