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 $(x + 1 5 \,t)$ where $x$ and $Y$ are measured in metre and time in second. The tension force in the string is

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

Answer 1

Answer: (C)

Given that, the linear mass density,
$m=10^{-3} kg / m$ and equation of the wave
$y=0.05 \sin (x+15 t)$...(i)
Since, the general equation of wave,
$y=a \sin (k x+\omega t)$...(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, $v=f \lambda=2 \pi \times \frac{15}{2 \pi}=15 m / s$
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
$T=(15)^{2} \times 10^{-3}=0.225 N$
Hence, the tension force in the string is $0.225 N$.