The vibrations of a string of length 60cm fixed at both ends are represented by the equation $y =…
The vibrations of a string of length 60cm fixed at both ends are represented by the equation $y = 4\sin\left(\frac{\pi x}{15}\right)\cos(96\pi t)$, where x and y are in cm and t in second.
(i) What is the maximum displacement of a point at $x = 5$ cm and $t = 0$?
(ii) Where are the nodes located along the string?
(iii) What is the velocity of the particle at $x = 7.5$ cm and at $t = 0.25$ s?
(iv) Write down the equations of the component waves whose superposition gives the above wave.
Solution
Sol. Given, $y = 4\sin\left(\frac{\pi x}{15}\right)\cos(96\pi t)$. It can be broken up into
$y = 2\left[\sin\left(\frac{\pi x}{15} + 96\pi t\right) + \sin\left(\frac{\pi x}{15} - 96\pi t\right)\right]$
Thus, the waves are of the same amplitude and frequency but travelling in opposite directions which thus, superimpose to give a standing wave.
(i) At $x = 5$ cm, the standing wave equation gives
$y = 4\sin\left(\frac{5\pi}{15}\right)\cos(96\pi t) = 4 \times \frac{\sqrt{3}}{2}\cos(96\pi t)$
$\therefore$ Maximum displacement = $2\sqrt{3}$ cm
(ii) The nodes are the points which are permanently at rest. Thus, they are those points for which
$\sin\left(\frac{\pi x}{15}\right) = 0$
i.e.
$\frac{\pi x}{15} = n\pi$
$\Rightarrow n = 0, 1, 2, 3, 4, \ldots$
$\Rightarrow x = 15n$, i.e. at $x = 0, 15, 30, 45$ and $60$ cm.
(iii) The particle velocity is equal to
$\displaystyle \frac{\partial y}{\partial t} = 4(96\pi)\sin\left(\frac{\pi x}{15}\right)(-\sin 96\pi t)$
$= -384\pi \sin\left(\frac{\pi x}{15}\right)\sin(96\pi t)$
At $x = 7.5$ and $t = 0.25$, we get
$\displaystyle \frac{\partial y}{\partial t} = -384\pi \sin\left(\frac{\pi \cdot 7.5}{15}\right)\sin(96\pi \cdot 0.25)$
$= -384\pi \sin\left(\frac{\pi}{2}\right)\sin(24\pi) = 0$
(iv) The equations of the component waves are
$y_1 = 2\sin\left(\frac{\pi x}{15} + 96\pi t\right)$
and
$y_2 = 2\sin\left(\frac{\pi x}{15} - 96\pi t\right)$