Question

A transverse sinusoidal wave on a string has a period T 5 25.0 ms and travels in the negative x direction with a speed of 30.0 m/s. At t 5 0, an element of the string at x 5 0 has a transverse position of 2.00 cm and is traveling downward with a speed of 2.00 m/s. (a) What is the amplitude of the wave?

Answer 1

A = 0.021525 m

Step-by-step explanation:

given,

Time period of string = T = 25 ms

velocity is -ve x-direction = 30 m/s

at t = 0 x = 0

transverse position = x = 2 cm = 0.02 m

speed = -2 m/s

Amplitude of wave = ?

At x = 0

x = A cos (ωt + ∅) at t= 0

x = A cos ∅

0.02 = A cos ∅.................(1)

v = - A ω sin(ωt + ∅) at t= 0

v = - A ω sin ∅

-2 = -A ω sin ∅

angular frequency

`\omega = (2\pi)/(T)`

`\omega = (2\pi)/(0.025)`

`\omega = 80\pi`

2 = A x 80 π sin ∅

`A sin \phi = (1)/(40\pi)`..............(2)

dividing equation (1) by (2)

`tan \phi = (1)/(40\pi * 0.02)`

`tan \phi = (1)/(0.8\pi)`

∅ = 0.3786

putting value in equation (1)

0.02 = A cos (0.3786)

A = 0.021525 m