Question

Sinusoidal waves $5.00cm$ in amplitude are to be transmitted along a string having a linear mass density equal to $4.00 \times {10^{ - 2}}kg{\text{ }}{m^{ - 1}}$ . If the source can deliver an average power of $90W$ and the string is under a tension of $100N$ , then find the frequency at which the source can operate is (take ${\pi ^2} = 10$)
A. $45Hz$
B. $50Hz$
C. $30Hz$
D. $62Hz$

Answer 1

Turning up the source, frequency will increase the power carried by the wave. Perhaps on the order of a hundred hertz will be the frequency at which the energy per second is $90J{\text{ }}{s^{ - 1}}$ . We will use the expression for power carried by a wave on a string.

Formula used:
$v = \sqrt {\dfrac{T}{\mu }}$
Where,
$v$ is the speed of wave,
$T$ is the tension and
$\mu$ is the mass per unit length.
$P = \dfrac{1}{2}\mu \upsilon {\omega ^2}{A^2}$
$P$ is the average power,
$\mu$ is the mass per unit length.
$\upsilon$ is the speed,
$\omega$ is the angular frequency,
$A$ is the amplitude.

Complete step by step solution:
Now we have to find the highest frequency. According to the question,
$A = 5cm = 0.05m$
$\mu = 0.04kg{\text{ }}{m^{ - 1}}$
$P = 90W$
$T = 100N$
Now,
$v = \sqrt {\dfrac{T}{\mu }} \\\ \Rightarrow v = \sqrt {\dfrac{{100}}{{0.04}}} \\\ \Rightarrow v = 50m{\text{ }}{s^{ - 1}} \\\$
Now, we will find, $\omega$
$\because P = \dfrac{1}{2}\mu \upsilon {\omega ^2}{A^2} \\\ \Rightarrow \omega = \sqrt {\dfrac{{2P}}{{\mu \upsilon {A^2}}}} \\\$
Now, substituting all the values,
${\omega ^2} = \dfrac{{2P}}{{\mu \upsilon {A^2}}} \\\ \Rightarrow {\omega ^2} = \dfrac{{2(90)}}{{(0.04)(50){{(0.05)}^2}}} \\\ \Rightarrow {\omega ^2} = 3.6 \times {10^4} \\\$
Now, as we know that the highest frequency at which the source can operate is given by
${\omega ^2} = {(2\pi f)^2} \\\ \Rightarrow 4{\pi ^2}{f^2} = 3.6 \times {10^4} \\\ \Rightarrow {f^2} = 900 \\\ \Rightarrow f = 30Hz \\\$
So, the frequency at which the source can operate is $30Hz$ .
Hence, the correct option is C.

Note:
This string wave would softly broadcast sound into the surrounding air, at the frequency of the second lowest note called A on piano. If we tried to turn the source to a higher frequency, it might just vibrate with smaller amplitude. The power is generally proportional to the squares of both the frequency and the amplitude.