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

Answer:

20.44 Hz

Explanation:

We are given that

Length of strand=L=11 m

Mass of strand=m=6 g=$6\times 10^{-3} kg$   (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=$v=\sqrt{\frac{T}{\mu}}$

Where T= Tension force

$\mu=\frac{m}{l}$

$\mu=\frac{6\times 10^{-3}}{11}=0.55\times 10^{-3}kg/m$

$v=\sqrt{\frac{6\times 9.8}{0.55\times 10^{-3}}}=326.97 m/s$

Frequency=$\frac{v}{2d}$

Substitute the values in the formula

Then, we get

Frequency=$\frac{326.97}{2\times 8}=20.44Hz$

Hence, the fundamental frequency of ist vibration=20.44 Hz