Question

A harmonic wave travels in the positive x direction at 12 m/s along a taught string. A fixed point on the string oscillates as a function of time according to the equationy = 0.0205 cos(4t)where y is the displacement in meters and the time t is in seconds(a) What is the amplitude of the wave, in meters?(b) What is the frequency of the wave, in hertz?(c) What is the wavelength of the wave, in meters?

Answer 1

Amplitude, A = 0.0205 m

Frequency, f = 0.636 Hz

Wavelength, 18.84 m

Explanation:

Given that,

Speed of the wave along a taught string, v = 12 m/s

A fixed point on the string oscillates as a function of time according to the equation :

`y = 0.0205\ \cos(4t)`........(1)

Here, y is the displacement in meters and the time t is in seconds

The general equation of oscillation is given by :

`y=A\ \cos\omega t`............(2)

On comparing equation (1) and (2) we get :

(a) Amplitude, A = 0.0205 m

(b) Frequency of the wave,

`\omega=4\\\\2\pi f=4\\\\f=(4)/(2\pi)\\\\f=0.636\ Hz`

(c) Velocity of the wave is given by formula as :

`v=(\omega)/(k)`

Since,

`k=(2\pi)/(\lambda)`

`v=(\omega\lambda)/(2\pi)\\\\\lambda=(2\pi v)/(\omega)\\\\\lambda=(2\pi * 12)/(4)\\\\\lambda=18.84\ m`

Hence, this is the required solution.