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

Answer:

A = 0.021525 m

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

$\omega = \dfrac{2\pi}{T}$

$\omega = \dfrac{2\pi}{0.025}$

$\omega = 80\pi$

2 = A x 80 π sin ∅

$A sin \phi = \dfrac{1}{40\pi}$..............(2)

dividing equation (1) by (2)

$tan \phi = \dfrac{1}{40\pi \times 0.02}$

$tan \phi = \dfrac{1}{0.8\pi}$

  ∅ = 0.3786

putting value in equation (1)

0.02  = A cos (0.3786)

A = 0.021525 m