Question

he total length of the strand is L = 11.0 m, the mass of the strand is m = 6.00 g, the mass of the hanging object is M = 6.00 kg, and the pulley is a fixed a distance d = 8.00 m from the wall. You pluck the strand between the wall and the pulley and it starts to vibrate. What is the fundamental frequency (in Hz) of its vibration?

Answer 1

20.44 Hz

Step-by-step explanation:

We are given that

Length of strand=L=11 m

Mass of strand=m=6 g=
`6* 10^(-3) kg` (1kg=1000g)

Mass of hanging object=M=6 kg

Distance of pulley from the wall=d=8 m

We have to find the fundamental frequency of(Hz) of its vibration.

Velocity=
`v=\sqrt{(T)/(\mu)}`

Where T= Tension force

`\mu=(m)/(l)`

`\mu=(6* 10^(-3))/(11)=0.55* 10^(-3)kg/m`

`v=\sqrt{(6* 9.8)/(0.55* 10^(-3))}=326.97 m/s`

Frequency=
`(v)/(2d)`

Substitute the values in the formula

Then, we get

Frequency=
`(326.97)/(2* 8)=20.44Hz`

Hence, the fundamental frequency of ist vibration=20.44 Hz