Question

A sinusoidal transverse wave travels along a long, stretched, string. the amplitude of this wave is 0.0885 m, it's frequency is 4.31 hz, and its wavelength is 1.21m.

(a) What is the transverse distance between a maximum and a minimum of the wave?
(b) How much time is required for 71.7 cycles of the wave to pass a stationary observer?
(c) Viewing the whole wave at any instant, how many cycles are there in a 30.7-m length of string?

Answer 1

(a) 0.177 m

(b) 16.491 s

(c) 25 cycles

Step-by-step explanation:

(a)

Distance between the maximum and the minimum of the wave = 2A ............ Equation 1

Where A = amplitude of the wave.

Given: A = 0.0885 m,

Distance between the maximum and the minimum of the wave = (2×0.0885) m

Distance between the maximum and the minimum of the wave = 0.177 m.

(b)

T = 1/f ...................... Equation 2.

Where T = period, f = frequency.

Given: f = 4.31 Hz

T = 1/4.31

T = 0.23 s.

If 1 cycle pass through the stationary observer for 0.23 s.

Then, 71.7 cycles will pass through the stationary observer for (0.23×71.7) s.

= 16.491 s.

(c)

If 1.21 m contains 1 cycle,

Then, 30.7 m will contain (30.7×1)/1.21

= 25.37 cycles

Approximately 25 cycles.