Question

One end of a string is attached to a rigid wall on a tabletop. The string is run over a frictionless pulley and the other end of the string is attached to a stationary hanging mass. The distance between the wall and the pulley is 0.405 meters, When the mass on the hook is 25.4 kg, the horizontal portion of the string oscillates with a fundamental frequency of 261.6 Hz (the same frequency as the middle C note on a piano). Calculate the linear mass density of the string.

Answer 1

0.005550 Kg/m

Step-by-step explanation:

The picture attached below shows the full explanation

Answer 2

The linear mass density is of the string
`\mu= 5.51*10^(-3) kg / m`

Step-by-step explanation:

From the question we are told that

The distance between wall and pulley is
`d = 0.405m`

The mass on the hook is
`m = 25.4\ kg`

The frequency of oscillation is
`f = 261.6 Hz`

Generally, the frequency of oscillation is mathematically represented as

`f = (1)/(2d) \sqrt{(T)/(\mu) }`

Where T is the tension mathematically represented as

T = mg

Substituting values

`T = 25.4 *9.8`

`=248.92N`

`\mu` is the mass linear density

Making
`\mu` the subject of the formula above

`\mu = (T)/((2df)^2)`

Substituting values

`\mu = (248.92)/((2 * 0.405 * 261.6)^2)`

`\mu= 5.51*10^(-3) kg / m`