Name: Condition for Resonance
Uploaded: 2019-07-19
Duration: 5 min 17 s
Description: Do you know you can shatter the glass goblet without touching it? Yes, it is possible when the sound of a specific frequency is produced. Let's find out the...

Question

A string of mass $m$ is fixed at both ends. The fundamental tone oscillations are excited with circular frequency $ω$ and maximum displacement amplitude $a_{\max}$ . Find:

(a) the maximum kinetic energy of the string;

(b) the mean kinetic energy of the string averaged over one oscillation period.

Accepted Answer

Let the two waves $y_{1} = a \sin (ω t + k x)$ incident wave and $y_{2} = a \sin (ω t - k x + π)$ reflected wave at fixed end superpose.

From the principle of superposition,

$y = y_{1} + y_{2}$

$\Rightarrow y = a \sin (ω t + k x) + a \sin (ω t - k x + π)$

$y = a \sin (ω t + k x) - a \sin (ω t - k x)$

Using the trigonometric formula

we have a standing wave (the resultant wave) in the string of the form, $y = 2 a \sin (k x) \cos (ω t)$ .

According to the problem, $m$ .

Hence, the standing wave excited in the string is,

$y = a_{m} \sin (k x) \cos (ω t)$ $. . . (1)$

From partial differential with respect to time, $v = \frac{\partial y}{\partial t} = - ω a_{m} \sin (k x) \sin (ω t)$ $. . . (2)$

So, the kinetic energy confined in the string element of length $d x$ , is given by,

$d K = \frac{1}{2} (d m) v^{2} = \frac{1}{2} (\frac{m}{l} d x) {(\frac{\partial y}{\partial t})}^{2}$

because particle velocity is product of wave velocity and slope of the string at that point.

or, $d K = \frac{1}{2} (\frac{m}{l} d x) a_{m}^{2} ω^{2} \sin^{2} (k x) \sin^{2} (ω t)$

or, $d K = \frac{m a_{m}^{2} ω^{2}}{2 l} \sin^{2} (ω t) \sin^{2} (\frac{2 π}{λ} x) d x$

Hence, the kinetic energy confined in the string, corresponding to the fundamental tone,

$K = \int d K = \frac{m a_{m}^{2} ω^{2}}{2 l} \sin^{2} (ω t) \int_{0}^{λ / 2} \sin^{2} (\frac{2 π}{λ} x) d x$

Because, for the fundamental mode, length of the string, $l = \frac{λ}{2}$

On integrating, we get, $K = \frac{1}{4} m a_{m}^{2} ω^{2} \sin^{2} (ω t)$

Hence, the sought maximum kinetic energy equals, $T_{\max} = \frac{1}{4} m a_{m}^{2} ω^{2}$ , because for $T_{\max}, \sin^{2} (ω t) = 1$

(ii) Mean kinetic energy averaged over one oscillation period,

$< K E > = \frac{\int K d t}{\int d t} = \frac{1}{4} m a_{m}^{2} ω^{2} \frac{\int_{0}^{2 π ω} \sin^{2} (ω t) d t}{\int_{0}^{2 π ω} d t}$

or, $< K E > = \frac{1}{8} m a_{m}^{2} ω^{2}$

A string of mass $m$ is fixed at both ends. The fundamental tone oscillations are excited with circular frequency $ω$ and maximum displacement amplitude $a_{\max}$ . Find:

(a) the maximum kinetic energy of the string;

(b) the mean kinetic energy of the string averaged over one oscillation period.

Important Questions on OSCILLATIONS AND WAVES

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A string of mass m is fixed at both ends. The fundamental tone oscillations are excited with circular frequency ω and maximum displacement amplitude amax. Find: (a) the maximum kinetic energy of the string; (b) the mean kinetic energy of the string averaged over one oscillation period.

Important Questions on OSCILLATIONS AND WAVES

Learn and Prepare for any exam you want !