Question

The linear density of a vibrating string is `1.3 xx 10^(-4) kg//m` A transverse wave is propagating on the string and is described by the equation `y= 0.021 sin (x + 30 t)` where x and y are measured in meter and t`t` in second the tension in the string is :-
A. 0.2 N
B. 0.250 N
C. 0.225 N
D. 0.325 N

Answer 1

Correct Answer - C
Given that, the linear mass density,
`m=10^(-3)kg//m` and equation of the wave
y = 0.05 sin(x + 15t) ….(i)
Since , the general equation of wave,
`y=asin(kx+omegat)` ….(ii)
Now, compairing the Eqs. (i) and (ii) we get,
`k=1,lamda=2pi(becausek=(2pi)/lamda)`
and `omega=15 rArrf=15/(2pi)` (`becauseomega=2pif`)
Velocity of the wave `v=flamda=2pixx15/(2pi)=15` m/s
As, we know, the tension force in the string,
`T=v^(2)m` `(becausev=sqrt(T/m))`
So, by substituting the values in the above relation, we get
`T=(15)^(2)xx10^(-3)=0.225` N
Hence, the tension force in the string is 0.225 N.