Question

The linear density of a vibrating spring is $1.3 \times {10^{ - 4}}kg/m$ . A transverse wave propagating on the string is described by the equation $y = 0.021\sin (x + 30t)$ where x and y are measured in meter and t in second. Tension in the string is
(A) \[0.12N\]
(B) $0.48N$
(C) $1.20N$
(D) $4.80N$

Answer 1

Hint: If the wavelength and the angular frequency of a wave is respectively $\lambda $ and $\omega $ then the equation of a wave is $y(x,t) = A\sin (\dfrac{{2\pi }}{\lambda }x \mp \omega t + \varphi )$ where A is the amplitude and $\varphi $ is the phase shift.

Complete step by step answer:
The velocity of a wave (V) = wavelength ($\lambda $) $ \times $ frequency (f)
Now the given wave equation is $y = 0.021\sin (x + 30t)$ where x and y are measured in meter and t in second. So comparing this equation with the general wave equation we get
$\dfrac{{2\pi }}{\lambda } = 1$ and $\omega = 30$
So $\lambda = 2\pi $
Now the angular frequency ($\omega $) is $30$ so the regular frequency (f) will be $\dfrac{{30}}{{2\pi }}$
The velocity of the wave is $ = \lambda \times f$ $ = 2\pi \times \dfrac{{30}}{{2\pi }} = 30$ $m/s$
We know that if the linear mass density and the tension of a string are respectively $\mu $ and $T$ the velocity of the wave propagating on the string is $V = \sqrt {\dfrac{T}{\mu }} $
According to the question $\mu = 1.3 \times {10^{ - 4}}kg/m$ and we got $V = 30m/s$
So, $T = {V^2} \times \mu $
$
   \Rightarrow T = 90 \times 1.3 \times {10^{ - 4}} \\
   \Rightarrow T = .117 \\
 $
Therefore, the tension, $T = .117N \approx 0.12N$

The correct answer is option A.

Note: The velocity of a wave propagating on a string can be got by the ratio of angular frequency and the wave number. In this particular problem the wave number (k) is $1$ and the relation will be $V = \dfrac{\omega }{k}$.