Question 17: Waves

Question

The vibrations of a string of length 60 cm fixed at both ends are represented by the equation
$$y = 4\sin \left( {{{\pi x} \over {15}}} \right)\cos (96\pi t)$$
where x and y are in cm and t is in second. Calculate the velocity of the particle at x = 7.5 cm at t = 0.25 s.

Accepted Answer

The given equation is
$$y = 4\sin \left( {{{\pi x} \over {15}}} ight)\cos (96\pi t)$$ ...... (i)
This equation is similar to the displacement equation of a travelling wave, $$y = A\sin (\omega t - kx)$$
On comparing these two equations, we get
$$\omega  = 2\pi v = 96\pi $$
or, $$
u  = {{96\pi } \over {2\pi }} = 48$$ Hz
Now, the velocity of the string at a point x at time t is obtained by differentiating Eq. (i) w.r.t.
$$	herefore$$ $$v = {{dy} \over {dt}} =  - (4 	imes 9\pi ) 	imes \sin \left( {{{\pi x} \over {15}}} ight)\sin (96\pi t)$$
At x = 7.5 cm = 7.5 $$	imes$$ 10
$$-$$2
m and t = 0.25 s,
$$v =  - 384\pi \sin \left( {\pi  	imes {{7.5 	imes {{10}^{ - 2}}} \over {15}}} ight)\sin \left( {96\pi  	imes {{25} \over {100}}} ight)$$
$$ =  - 384\pi \sin \left( {{\pi  \over {200}}} ight)\sin (24\pi )$$
$$ = 0$$ ($$\because$$ $$\sin 24\pi  = 0$$)
Hence, the velocity of the particle is zero.

Answer

4 m/s

Answer

0

Answer

16 m/s

Answer

9.8 m/s

Waves

The vibrations of a string of length 60 cm fixed at both ends are represented by the equation $y = 4\sin \left( {{{\pi x} \over {15}}} \right)\cos (96\pi t)$ where x and y are in cm and t is in second. Calculate the velocity of the particle at x = 7.5 cm at t = 0.25 s.

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