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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?

User Domruf
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1 Answer

2 votes

Answer:

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

User The Paul
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