Question 468: Waves

Question

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

Accepted Answer

Given that, the linear mass density,
$m=10^{-3} \mathrm{~kg} / \mathrm{m}$ and equation of the wave
$$y=0.05 \sin (x+15 t)\quad 	ext{.... (i)}$$
Since, the general equation of wave,
$$y=a \sin (k x+\omega t)\quad 	ext{... (ii)}$$
Now, compairing the Eqs. (i) and (ii) we get,
$$k=1, \lambda=2 \pi \quad\left(\because k=\frac{2 \pi}{\lambda}ight)$$
and $\omega=15 \Rightarrow f=\frac{15}{2 \pi} (\because \omega=2 \pi f)$
Velocity of the wave, $v=f \lambda=2 \pi 	imes \frac{15}{2 \pi}=15 \mathrm{~m} / \mathrm{s}$
As, we know, the tension force in the string,
$$T=v^2 m \quad\left(\because v=\sqrt{\frac{T}{m}}ight)$$
So, by substituting the values in the above relation, we get
$$T=(15)^2 	imes 10^{-3}=0.225 \mathrm{~N}$$
Hence, the tension force in the string is 0.225 N.

Answer

0.2 N

Answer

0.250 N

Answer

0.225 N

Answer

0.325 N

Waves

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

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