Q.

Shape of string transmitting wave along x-axis at some instant is shown. Velocity of point P is v=4π cm/s and θ=tan−10.004π

Moderate

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By Expert Faculty of Sri Chaitanya

Amplitude of wave is 2 mm

Velocity of wave is 10 m/s

Max acceleration of particle is 80π2 cm/sec2

Wave is traveling in – ve x- direction

answer is 1,2,3,4.

(Detailed Solution Below)

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Detailed Solution

Wave length of the wave, λ=1m maximum particle velocity, Vmax=Aω=4π ∴wave velocity, v=∂y/∂t∂y/∂x=4π×10−2tanθ=4π×10−24π×10−3=10m/s ∵wave velocity=particle velocityslope of the wave also frequency, f or ϑ=Vλ=V1=10HzVmax=Aω4π×10−2=A2π(10)Amplitude of the waveA=2×10−3m and maximum accelertion of the particle amax=Aω2=80π2cm/sec2

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