Question

A transverse wave is propagating on the string.The linear density of a vibrating string is $10^{-3}kg/m$. The equation of the wave is $Y=0.05$ sin $(x+15t)$ where $x$ and $Y$ are measured in metre and time in second. The tension force in the string is

• A. $0.2N$
• B. $0.250N$
• C. $0.225N$
• D. $0.325N$

Answer 1

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,
$y=a\sin (kx+\omega t)$...(ii)
Now, compairing the Eqs. (i) and (ii) we get,
$k=1,\lambda =2\pi \left(\because k=\frac{2\pi }{\lambda }\right)$
and $\omega =15\Rightarrow f=\frac{15}{2\pi }(\because \omega =2\pi f)$
Velocity of the wave, $v=f\lambda =2\pi \times \frac{15}{2\pi }=15m/s$
As, we know, the tension force in the string,
$T=v^{2}m\left(\because v=\sqrt{\frac{T}{m}}\right)$
So, by substituting the values in the above relation, we get
$T=(15{)}^2\times 10^{-3}=0.225N$
Hence, the tension force in the string is $0.225N$.