Concepts Of Physics MCQ Edition [Volume 1]

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Wave Motion and Waves on a String

A string clamped at both ends is set into vibration by a 660 Hz...

A string clamped at both ends is set into vibration by a

660 \text{ Hz}

tuning fork, causing it to vibrate in three loops. The wave speed for a transverse wave on the string is

220 \text{ m s}^{-1}

. If the maximum amplitude of a particle is

0.5 \text{ cm}

, which of the following is a suitable equation describing the motion?

Options

A $(0.5 \text{ cm}) \sin[(0.06\pi \text{ cm}^{-1})x] \cos[(1320\pi \text{ s}^{-1})t]$
B $(0.5 \text{ cm}) \sin[(0.06\pi \text{ cm}^{-1})x] \cos[(660\pi \text{ s}^{-1})t]$
C $(0.5 \text{ cm}) \cos[(0.06\pi \text{ cm}^{-1})x] \cos[(1320\pi \text{ s}^{-1})t]$
D $(0.5 \text{ cm}) \sin[(6\pi \text{ cm}^{-1})x] \cos[(1320\pi \text{ s}^{-1})t]$

A. $(0.5 \text{ cm}) \sin[(0.06\pi \text{ cm}^{-1})x] \cos[(1320\pi \text{ s}^{-1})t]$

Given frequency

f = 660 \text{ Hz}

and wave speed

v = 220 \text{ m s}^{-1} = 22000 \text{ cm s}^{-1}

.

Angular frequency

\omega = 2\pi f = 2\pi \times 660 = 1320\pi \text{ rad s}^{-1}

.

Wave number

k = \dfrac{\omega}{v} = \dfrac{1320\pi}{22000} = 0.06\pi \text{ cm}^{-1}

.

Since the string is clamped at both ends, the displacement at

x = 0

must be zero at all times. Therefore, the spatial part of the standing wave equation must be a sine function,

\sin(kx)

.

The maximum amplitude of the standing wave is given as

A = 0.5 \text{ cm}

.

The equation of the standing wave is of the form

y(x,t) = A \sin(kx) \cos(\omega t)

.

Substituting the values, we get:

y(x,t) = (0.5 \text{ cm}) \sin[(0.06\pi \text{ cm}^{-1})x] \cos[(1320\pi \text{ s}^{-1})t]