Question

A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is

0.0901

m, its frequency is

2.27

Hz, and its wavelength is

1.05

m.

(a) What is the shortest transverse distance between a maximum and a minimum of the wave?

shortest transverse distance:

m

(b) How much time is required for

57.9

cycles of the wave to pass a stationary observer?

time to pass a stationary observer:

s

(c) Viewing the whole wave at any instant, how many cycles are there in a

30.3

m length of string?

cycles

Answer 1

Part A.

In a complete wave shortest distance between maximum and minimum of wave = wavelength/2

given that wavelength = 1.05 m

shortest transverse distance = 1.05/2 = 0.525 m

Part B.

Given that frequency of wave = 2.27 Hz = 2.27 cycles/sec

So time required for 57.9 cycles will be:

t = N/f = 57.9 cycles/(2.27 cycles/sec) = 57.9/2.27

t = 25.5 sec

Part C.

Since we know that wavelength = distance between two maximum, So

wavelength = 1.05 m/cycle

Length of string = 30.3 m

So,

N = 30.3/(1.05 m/cycle) = 28.86 cycles