Question

The linear density of a vibrating string is $ 1.3 \times {10^{ - 4}}{{kg} \mathord{\left/ {\vphantom {{kg} m}} \right.} m} $ . A transverse wave is propagating on the string and is described by the equation $ y = 0.021\sin \left( {x + 30t} \right) $ where $ x $ and $ y $ are measured in meters and $ t $ in seconds. What is the tension in the string?
(A) $ 0.12N $
(B) $ 0.48N $
(C) $ 1.20N $
(D) $ 4.80N $

Answer 1

Hint : We need to use the general equation of wave to derive the relation of velocity to angular frequency. Then we can find the relationship of velocity with tension and finally we can equate the value of v.

Formula Used: The following formulae are used to solve this question,
 $\Rightarrow v = \sqrt {\dfrac{T}{\rho }} $ where, $ v $ is the velocity of a wave, $ T $ is the tension and $ \rho $ is the linear density.
General equation of a wave: $ y = A\sin (kx + \omega t) $ where $ k $ is the wave number, and $ \omega $ is the angular frequency of the wave.
 $\Rightarrow \omega {\text{ }} = {\text{ }}2\pi /t{\text{ }} = {\text{ }}2\pi f $ where $ t $ is the time period and $ f $ is the frequency.

Complete step by step answer
It is given that 1m length of the string weighs $ 1.3 \times {10^{ - 4}}kg $ .
That is, the linear density of the string is $ \rho = 1.3 \times {10^{ - 4}}{{kg} \mathord{\left/
 {\vphantom {{kg} m}} \right.} m} $ .
Now, a transverse wave is one where the displacement of particles is perpendicular to the direction of propagation of waves.
For a transverse harmonic wave travelling in the negative $ x $ -direction we have
$\Rightarrow \;y\left( {x,t} \right) = Asin\left( {kx + \omega t} \right) = Asin\left( k \right.\left. {\left( {x + vt} \right)} \right){\text{ }} $
 $ \therefore $ The velocity v of a wave can be given by-
 $\Rightarrow v = \dfrac{\omega }{k} $
It is given in the question, $ y = 0.021\sin \left( {x + 30t} \right) $ .
 $ \therefore v = \dfrac{\omega }{k} = \dfrac{{30}}{1}{m \mathord{\left/
 {\vphantom {m {s.}}} \right.} {s.}} $
 $ \Rightarrow v = 30{m \mathord{\left/
 {\vphantom {m s}} \right.} s} $
Now, $ v = \sqrt {\dfrac{T}{\rho }} $ where, $ v $ is the velocity of a wave, $ T $ is the tension and $ \rho $ is the linear density.
It is given, the linear density $ \rho = 1.3 \times {10^{ - 4}}{{kg} \mathord{\left/
 {\vphantom {{kg} m}} \right.} m} $ and $ v = 30{m \mathord{\left/
 {\vphantom {m s}} \right.} s} $ .
Therefore we get on substituting, $ 30 = \sqrt {\dfrac{T}{{1.3 \times {{10}^{ - 4}}}}} $
Squaring on both sides we get,
 $ \Rightarrow 900 \times 1.3 \times {10^{ - 4}} = T $
The unit of $ T $ is $ {\text{kg m }}{{\text{s}}^{{\text{ - 2}}}} $ and $ {\text{1N = 1kg m }}{{\text{s}}^{{\text{ - 2}}}} $
 $ \therefore $ $ T = 0.117N \approx 0.120N $
The correct answer is Option A.

Note
If we consider a transverse wave travelling in the positive $ x $ -direction. The displacement $ y $ of a particle in the medium is given as a function of $ x $ and $ t $ by the general wave equation-
 $ y = A\sin (kx + \omega t) $ where $ k $ is the wave number, $ t $ is time in seconds, $ x,y $ are displacements in the respective directions and $ \omega $ is the angular frequency of the wave.