# One End of a Long String of Linear Mass Density 8.0 × 10–3 Kg M–1 Is Connected to an Electrically Driven Tuning Fork of Frequency 256 Hz. 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 - Physics

One end of a long string of linear mass density 8.0 × 10–3 kg 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 = 0, the left end (fork end) of the string = 0 has zero transverse displacement (= 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement as the function of and that describes the wave on the string.

#### Solution 1

The equation of a travelling wave propagating along the positive y-direction is given by the displacement equation:

y (x, t) = a sin (wt – kx) … (i)

Linear mass density,  mu = 8.0 xx 10^(-3) kg m^(-1)

Frequency of the tuning fork, ν = 256 Hz

Amplitude of the wave, = 5.0 cm = 0.05 m … (ii)

Mass of the pan, = 90 kg

Tension in the string, T = mg = 90 × 9.8 = 882 N

The velocity of the transverse wave v, is given by the relation:

v = sqrt(T/mu)

= sqrt(882/(8.0xx10^(-3))) = 332 "m/s"

Angular frequency , omega = 2piv

= 2xx 3.14xx 256

= 1608.5 = 1.6 xx 10^(3) "rad/s"  ....(iii)

Wavelength, lambda = v/v = 332/256 m

:. Propagation constant, k = (2pi)/lambda

= (2xx3.14)/(332/256) = 4.84 m^(-1)  ....(iv)

Substituting the values from equations (ii), (iii), and (iv) in equation (i), we get the displacement equation:

(xt) = 0.05 sin (1.6 × 103t – 4.84 x) m

#### Solution 2

Here, mass per unit length, g = linear mass density = 8 x 10-3 kg m-1;
Tension in the string, T = 90 kg = 90 x 9.8 N= 882 N;

Frequency v = 256 Hz and amplitude A = 5.0 cm = 0.05 m

As the wave propagating along the string isa transverse travelling wave, the velocity of the wave

v = sqrt(T/mu) = sqrt(882/8xx10^(-3)) ms^(-1) = 3.32 xx 10^2 ms^(-1)

Now omega = 2piv = 2 xx 3.142 xx 256 = 1.61 xx 10^2 ms^(-1)

Also v = vlambda or lambda = v/v  = (3.32 xx 10^2)/256 m

Propagation constant, k = (2pi)/lambda = (2xx3.142xx256)/(3.32xx10^2)

= 4.84 m^(-1)

:. The equation of of the wave is

v(x,t) = A sin(omegat - kx)

= 0.05 sin (1.61 xx 10^3 t - 4.84 x)

Here x,y are in meter  and t is in second

Concept: Displacement Relation for a Progressive Wave
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#### APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 15 Waves
Q 24 | Page 389