Question

A sinusoidal wave with wavelength 0.500 m travels along a string. The maximum transverse speed of a point on the string is. 4.00 m/s and the maximum transverse acceleration is 1.00 x 105 m/s2. What is the propagation speed of the wave

Answer 1

The velocity
`v = 1989.2 \ m/s`

Step-by-step explanation:

From the question we are told that

The wavelength is
`\lambda = 0.500 \ m`

The maximum transverse speed is
`v = 4.0 \ m/s`

The maximum transverse acceleration is
`a = 1.00 *10^(5) \ m/s^2`

Generally the frequency of the wave is mathematically represented as

`f = (w)/(2 \pi )`

Here w is the angular speed which is mathematically evaluated as

`w = (a)/(v)`

=>
`w = (1.00 *10^(5))/(4)`

=>
`w = 25000 \ rad/sec`

So

`f = ( 25000 )/(2 * 3.142 )`

=>
`f = ( 25000 )/(2 * 3.142 )`

=>
`f = 3978.4 \ Hz`

Gnerally the propagation speed of the wave is mathematically represented as

`v = f * \lambda`

=>
`v = 3978.4 * 0.500`

=>
`v = 1989.2 \ m/s`