A string of mass 0.2 kg/m and length l= 0.6 m is fixed at both ends and streteched such that it has a tension of 80 N. the string vibrates in 3 segments with maximum amplitude of 0.5 cm. the maximum transevers velcotiy amplitude is

Options

A. `9.43 ms^(-1)`
B. `3.14 ms^(-1)`
C. `1.57 ms^(-1)`
D. `6.28 ms^(-1)`

Answer

c

Explanation

As the string is vibrating in three segments, Therefore,
`l=(3lambda)/2 rArr lambda = (2l)/3=(2(0.6))/3=0.4` m
Now, `v=sqrt(T/m) rArr v= sqrt(80/2.0)=20 ms^(-1)`
`n= v/lambda = 20/0.4 = 50` Hz
Amplitude of particle belocity `=(dy/dt)_(max)=(a_(max))omega="a"_(max )(2pin)`
`=(0.5xx10^(-2))xx2pixx50=1.57 ms^(-1)`