Waves

Question : 24

Total: 27

One end of a long string of linear mass density 8.0×10–3‌kg‌m–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=8.0×10–3‌kg‌m–1, υ=256 Hz,
T = 90 kg = 90×9.8 = 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=√

=√

882

8.0×10−3

=3.32×102ms−1
ω=2πυ=2×

×256 =1.61×103‌rad‌s−1
λ=

3.32×102

256

m
Propagation constant,
k=

2π

2×3.142×256

3.32×102

=4.84‌rad‌m−1
As, the wave is propagating along positive x direction, the equation of the wave is
y(x,t)=A‌sin(ωt−kx)
=0.05‌sin(1.61×103t−4.84x)
Here, x, y are in metre and t is in second.

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