Question

The equation of a wave on a string of linear mass density $0.04\ \text{kg/m}$ is given by $y=0.02\left( m \right)\sin \left[ 2\pi \left( \dfrac{t}{0.04\left( s \right)}-\dfrac{x}{0.50\left( s \right)} \right) \right]$. The tension in the string is:
(A) 4.0 N
(B) 12.5 N
(C) 0.5 N
(D) 6.25 N

Answer 1

Hint: The standard wave equation for the string is represented with the following expression:
$y=A\sin \left( \omega t-kx \right)$
By comparing this standard equation with the given wave equation, we can easily get the values of $\omega $and K. After that we can easily find out the value of tension in the string.
Formula used:
The tension in the string is given by the following formula:
$T=\dfrac{\mu {{\omega }^{2}}}{{{k}^{2}}}$

Complete answer:
The equation of wave in a given string is $y=0.02\left( m \right)\sin \left[ 2\pi \left( \dfrac{t}{0.04\left( s \right)}-\dfrac{x}{0.50\left( s \right)} \right) \right]\cdots \cdots \left( 1 \right)$
Mass density = $0.04\ \text{kg/m}$
The standard wave equation for the string is represented with the following expression:
$y=A\sin \left( \omega t-kx \right)\cdots \cdots \left( 2 \right)$
By comparing equation (1) and (2) we get
$\omega =\dfrac{1}{0.04}\ \text{rad/sec}$
$k=\dfrac{1}{0.50}\ \text{rad/m}$
Now, the tension in the string is given by the following expression:
$T=\dfrac{\mu {{\omega }^{2}}}{{{k}^{2}}}$
Where
$\mu =$Mass density of the wave of string
Now substitute all the necessary values in the above formula we get
$T=\dfrac{\left( 0.04 \right){{\left( \dfrac{1}{0.04} \right)}^{2}}}{{{\left( \dfrac{1}{0.50} \right)}^{2}}}$
When we solve the above equation we get, $T=6.25N$
This is the value of tension in the string.

Therefore, option D is the correct answer.

Additional information:
The characteristics of the wave depends upon the medium in which the wave travels. The best example of this string wave production is guitar. In guitar the string wave produces the sound. The speed of the waves in the guitar determines the frequency of the sound. The strings are made of different materials so that they have different linear mass densities. Due to different mass density, they have different values on tensions produced in the string. The linear density is represented by the following expression:
$\mu =\dfrac{m}{L}$
Where
M= mass of the string
L = length of the string

Note:
In this problem we have to be careful while comparing the equation of string with the standard equation. After that carefully substitute the values in the given formula and then find the value of the tension in the string. Also, the ratio $\dfrac{\omega }{k}$ that we used in solving is actually the wave velocity. So the equation that we used to solve the problem can be rewritten as,
$T=\dfrac{\mu {{\omega }^{2}}}{{{k}^{2}}}=\mu {{v}^{2}}$
Where, v is the wave velocity.