Waves

Question : 24 of 27

Marks: +1, -0

One end of a long string of linear mass density

8.0 \times 10^{-3} \, \mathrm{kg} \, \mathrm{m}^{-1}

is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At

t = 0

, the left end (fork end) of the string

x = 0

has zero transverse displacement

(y = 0)

and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of

x

and

t

that describes the wave on the string.

Solution:

Here,

m = 8.0 \times 10^{-3} \, \mathrm{kg} \, \mathrm{m}^{-1}

\upsilon = 256

Hz,

T = 90 kg =

90 \times 9.8

= 882 N

Amplitude of wave, A = 5.0 cm = 0.05 m.

As, the wave propagating along the string is a transverse travelling wave, the velocity of the wave is given by

v = \sqrt{ \frac{T}{\mu} } = \sqrt{ \frac{882}{8.0 \times 10^{-3}} }

= 3.32 \times 10^{2} \, \mathrm{m} \, \mathrm{s}^{-1}

\omega = 2 \pi \upsilon = 2 \times \frac{22}{7} \times 256

= 1.61 \times 10^{3} \, \mathrm{rad} \, \mathrm{s}^{-1}

\lambda = \frac{v}{\upsilon} = \frac{3.32 \times 10^{2}}{256} \, \mathrm{m}

Propagation constant,

k = \frac{2 \pi}{\lambda} = \frac{2 \times 3.142 \times 256}{3.32 \times 10^{2}} = 4.84 \, \mathrm{rad} \, \mathrm{m}^{-1}

As, the wave is propagating along positive x direction, the equation of the wave is

y(x, t) = A \sin (\omega t - k x)

= 0.05 \sin (1.61 \times 10^{3} t - 4.84 x)

Here, x, y are in metre and t is in second.

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