Question

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 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.

Answer 1

Given, the linear mass density is 8.0× 10 −3   kg/m , the frequency of the tuning fork is 256 Hz, the mass of the tuning fork is 90 kg and the amplitude of the wave is 5.0 cm.

The tension in the string is given by the equation,

T=mg

Substituting the values in the above equation, we get:

T=90 kg×9.8 m/ s 2 =882 N

The velocity of the wave is given by the equation,

v= T μ

Substituting the values in the above expression, we get:

v= 882 N 8.0× 10 −3   kg/m =3.32× 10 2  m/s

The angular velocity is given by the equation,

ω=2πν

Substituting the values in above expression, we get:

ω=2π×256 Hz =1.61× 10 3   rad/s

The wavelength of the wave is given by the equation,

λ= v ν

The propagation constant is given by the equation,

k= 2π λ

Substituting the value in the above equation, we get:

k= 2π v ν = 2πν v

Substituting the values in the above equation, we get:

k= 2π( 256 Hz ) 3.32× 10 2  m/s =4.84  m −1

The general equation of a wave is,

y=Asin( ωt−kx )

Substituting the values in the above equation, we get:

y=0.05sin( 1.61× 10 3 t−4.84x )

Hence, the transverse displacement function is y=0.05sin( 1.61× 10 3 t−4.84x ).