Question

A tuning fork of frequency 440 Hz is attached to a long string of linear mass density \[0.01\,kg{m^{ - 1}}\] kept under a tension of 49 N. The fork produces transverse waves of amplitude 0.50 mm on the string.
(a) Find the wave speed and the wavelength of the waves.
(b) Find the maximum speed and acceleration of a particle of the string.
(c) At what average rate is the tuning fork transmitting energy to the string?

Answer 1

Hint: Transverse wave motion occurs when all points on a wave oscillate along pathways that are perpendicular to the wave's advance direction. Transverse waves include surface ripples on water, seismic waves, and electromagnetic (e.g., radio and light) waves. In this problem we are going to apply the expressions of wave equation and wave velocity.

Formula Used:
The formula to find velocity is,
\[v = \sqrt {\dfrac{T}{\mu }} \]
Where, T is tension in wire and \[\mu \] is mass per unit length.

Complete step by step solution:
Consider a tuning fork of frequency \[\nu = 440\,Hz\] that is attached to a long string of linear mass density \[\mu = 0.01\,kg{m^{ - 1}}\]kept under a tension \[T = 49N\]. The fork produces transverse waves of amplitude \[A = 0.50\,mm\] on the string.

(a) To find the wave speed and the wavelength of the waves we have,
\[v = \sqrt {\dfrac{T}{\mu }} \] ………. (1)
Substitute the value of T and \[\mu \]in equation (1) we get,
\[v = \sqrt {\dfrac{{49}}{{0.01}}} \]
\[\Rightarrow v = 70\,m{s^{ - 1}}\]
Also, \[v = f\lambda \]
\[\lambda = \dfrac{v}{f}\]
\[\Rightarrow \lambda = \dfrac{{70 \times {{10}^2}}}{{440}}\]
\[\Rightarrow \lambda = 15.9\,cm\]
\[ \therefore \lambda \simeq 16\,cm\]

Therefore, the wave speed is \[v = 70\,m{s^{ - 1}}\] and the wavelength of the waves is \[\lambda \simeq 16\,cm\].

(b) To find the maximum speed and acceleration of a particle of the string, we have
\[y = A\sin \left( {wt - kx} \right)\]
\[\Rightarrow v = \dfrac{{dy}}{{dt}} = A\omega \cos \left( {\omega t - kx} \right)\]
\[\Rightarrow {v_{\max }} = A\omega \]
\[\Rightarrow {v_{\max }} = A\left( {2\pi f} \right)\]
Now, substitute the value of f and A in above equation, we get
\[{v_{\max }} = 0.50 \times {10^{ - 3}}\left( {2 \times 3.142 \times 440} \right)\]
\[\Rightarrow {v_{\max }} = 1.381m{s^{ - 1}}\]
Now, the acceleration is,
\[a = \dfrac{{dv}}{{dt}}\]
\[\Rightarrow a = - A{\omega ^2}sin\left( {\omega t - kx} \right)\]
\[\Rightarrow {a_{\max }} = A{\omega ^2}\]
\[\Rightarrow {a_{\max }} = A{\left( {2\pi f} \right)^2}\]
\[\Rightarrow {a_{\max }} = 0.50 \times {10^{ - 3}}{\left( {2 \times 3.142 \times 440} \right)^2}\]
\[\therefore {a_{\max }} = 3.82\,km{s^{ - 2}}\]

Hence, the maximum speed is \[{v_{\max }} = 1.381\,m{s^{ - 1}}\] and acceleration of a particle of the string is \[{a_{\max }} = 3.82\,km{s^{ - 2}}\].

(c) To find at what average rate is the tuning fork transmitting energy to the string, we have
The average rate, P is given by,
\[P = 2{\pi ^2}{f^2}v{A^2}\]
Substitute the value of f, v and A we get,
\[P = 2 \times {\left( {3.142} \right)^2} \times \left( {440} \right) \times 70 \times {\left( {0.5 \times {{10}^{ - 3}}} \right)^2}\]
\[\therefore P = 0.67\,W\]

Hence, the average rate is the tuning fork transmitting energy to the string is \[P = 0.67W\].

Note: In this question it is important to remember the formulas for the velocity, acceleration, frequency, wavelength and the general equation of wave. Moreover, wave speed is the distance travelled by a wave in a particular length of time, such as the number of metres per second.