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Question

For a certain transverse standing wave on a long string, an antinode is formed at x=0 and next to it, a node is formed at x=0.10 m. The position y(t) of the string particle at x=0 is shown in figure.


A
Transverse displacement of the particle at x=0.05 m and t=0.05 s is 22 cm.
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B
Transverse displacement of the particle at x=0.04 m and t=0.025 s is 22 cm.
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C
Speed of the travelling waves that interfere to produce this standing wave is 2 m/s.
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D
The transverse velocity of the string particle at x=115 m and t=0.1 s is 20π cm/s.
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Solution

The correct options are
A Transverse displacement of the particle at x=0.05 m and t=0.05 s is 22 cm.
C Speed of the travelling waves that interfere to produce this standing wave is 2 m/s.
D The transverse velocity of the string particle at x=115 m and t=0.1 s is 20π cm/s.
Distance between an antinode and a node is equal to one fourth of the wavelength of the wave.
λ4=0.1λ=0.4m
From graph T=0.2 s
and amplitude of standing wave is 2A=4 cm.
Equation of the standing wave
y(x,t)=2Acos(2π0.4x)sin(2π0.2t)cmy(x=0.05,t=0.05)=22 cm
y(x=0.04,t=0.025)=22cos36
Speed =λT=2 m/s
Vy=dydt=2A×2π0.2cos(2πx0.4)cos(2πt0.2)Vy=(x=115m,t=0.1)=20π cm/s

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