Question

A string with linear mass density $m=5.00\times {10}^{-2}kg/m$ is under a tension of $80.0N.$ How much power must be applied to the string to generate sinusoidal waves at a frequency of $60.0\text{ Hz}$ and amplitude of $6.00\text{ cm}$
A.$512W$
B.$312W$
C.$614W$
D.$215W$

Answer 1

Hint: The relation to find the power supplied to the string is given by $p={\displaystyle \frac{1}{2}}\mu {\omega }^2{A}^{2}v$ using this relation try to analyze the terms that are present in the above relation and find out the power supplied by finding out the unknowns in the above relation with the given data.

Complete answer:
 We know that the relation to find the power supplied to the string is $p={\displaystyle \frac{1}{2}}m{\omega }^2{A}^{2}v$
Where $p$ is the power, $m$ is the mass density, $\omega$ is the angular frequency, is the amplitude of the wave and $v$ is the velocity
 We know that wave speed on the string is
$v=\sqrt{{\displaystyle \frac{T}{\mu }}}=\sqrt{{\displaystyle \frac{80}{0.05}}}=40m/s$
Given the frequency is $\vartheta =60Hz$
Therefore angular frequency is $\omega =2\pi f=2\pi (60)=377s^{-1}$
Given amplitude of the wave is $A=0.06m$
Also given that the mass density $m=5.00\times {10}^{-2}kg/m$
Put all these values in the above relation of the power
$p={\displaystyle \frac{1}{2}}\mu {\omega }^2{A}^{2}v={\displaystyle \frac{1}{2}}\times 0.05\times {377}^2\times {0.06}^2\times 40=511.66W$
Therefore the power required to supply for the string is $p=512W$

Hence the correct answer is option A.

Note:
While calculation make sure that all the terms involved in the formula are in the same system of units if not convert them in such a way that the whole terms are in the same system of units. And when there are some unknown terms in the formula we are using then try to find those terms using the given data and then put those values in the formula to find the required answer.