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=0.10m. The position y(t) of the string particle at x=0 is shown in figure.
A
Transverse displacement of the particle at x=0.05m and t=0.05s is −2√2cm.
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B
Transverse displacement of the particle at x=0.04m and t=0.025s is −2√2cm.
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C
Speed of the travelling waves that interfere to produce this standing wave is 2m/s.
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D
The transverse velocity of the string particle at x=115m and t=0.1s is 20πcm/s.
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Solution
The correct options are A Transverse displacement of the particle at x=0.05m and t=0.05s is −2√2cm. C Speed of the travelling waves that interfere to produce this standing wave is 2m/s. D The transverse velocity of the string particle at x=115m and t=0.1s is 20πcm/s. Distance between an antinode and a node is equal to one fourth of the wavelength of the wave. ∴λ4=0.1⇒λ=0.4m From graph ⇒T=0.2 s and amplitude of standing wave is 2A=4cm. Equation of the standing wave y(x,t)=−2Acos(2π0.4x)sin(2π0.2t)cmy(x=0.05,t=0.05)=−2√2cm y(x=0.04,t=0.025)=−2√2cos36∘ Speed =λT=2m/s Vy=dydt=−2A×2π0.2cos(2πx0.4)cos(2πt0.2)Vy=(x=115m,t=0.1)=20πcm/s