Question

The mass per unit length of a uniform wire is$0.135g/cm$. A transverse wave of the form $y=–0.21\sin (x+30t)$is produced in it, where $x$is in meter and$t$ is in second. Then, the expected value of tension in the wire is$x\times {10}^{-2}N$. Value of $x$is __________. (Round-off to the nearest integer)

Answer 1

Step 1. Given data:

1. The mass per unit length of a uniform wire$\frac{m}{l}=\mu =0.135g/cm$
2. A transverse wave of the form $y=–0.21\sin (x+30t)$
3. The expected value of tension in the wire is$x\times {10}^{-2}N$

Step 2. Formula used:

The speed of a pulse or wave on a string under tension can be written as,

$\left|v\right|=\sqrt{\frac{T}{\mu }}$………(1)

where$T$ is the tension in the string and$\mu$ is the mass per length of the string.

The phase velocity of a wave is the rate at which the wave packet propagates in any medium,

$v=\frac{\omega }{k}$ (where $\omega =$Angular frequency)…….(2)

Step 3. Calculating the value of$x$,

Now compare the given equation $y=–0.21\sin (x+30t)$to the general equation$y=A\sin (kx+\omega t)$.

$k=1,\omega =30$

Therefore the value of v from the equation of transverse wave,

Using the equation (2)-

$$\begin{gathered} v=\frac{\omega }{k} \\ v=\frac{30}{1} \\ v=30m/s \end{gathered}$$

Simplify further,

$\mu =0.135g/cm=0.0135kg/m$

Substituting the known values in the equation-

$\begin{array}{rcl}\left|v\right|& =& \sqrt{\frac{T}{\mu }}\\ 30& =& \sqrt{\frac{x\times {10}^{-2}}{0.0135}}\\ x& =& \frac{{30}^2\times 0.0135}{{10}^{-2}}\\ & =& 1215\end{array}$

Hence, the value of $x$is$1215$.