Question

A sinusoidal transverse wave travels along a long stretched string. The amplitude of this wave is 0.0817 m, its frequency is 3.61 Hz, and its wavelength is 1.11 m. (a) What is the transverse distance between a maximum and a minimum of the wave? (b) How much time is required for 56.9 cycles of the wave to pass a stationary observer? (c) Viewing the whole wave at any instant, how many cycles are there in a 34.1-m length of string?

Answer 1

a) The transverse distance is the perpendicular distance between two locations.

Amplitude is perpendicular distance of either maximum or minimum from mean position (initial position of the string). Hence the transverse distance between the maximum and minimum is twice the amplitude as maximum and minimum lies on either side of the mean position.

The transverse distance between a maximum and a minimum of the wave is 2A = 0.1634 m

b) The time taken by one cycle to pass through a point is time period T which is inverse of frequncy. T = 1 / f . Hence T = 1 / 3.61 s = 0.277 s

The time for 56.9 cycles of the wave to pass a stationary observer is t = 56.9 x 0.277 = 56.76 s

c) One cycle of wave indicates the wavelength. Hence the number of cycles in 34.1 m length of the string is n = 34.1 / wavelength = 34.1 / 1.11 = 30.72

The number of cycles in 34.1m length of string is 30.