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Q.

A particle  is executing simple harmonic motion along a straight line. When its distance from the equilibrium position is 3cm its velocity is 20cm/s. When the particle is at a distance of 4 cm from the equilibrium position, its velocity is 15cm/s . Then amplitude of oscillation is

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Detailed Solution

Using v=ωA2-x2,We can write 20=ωA2-32, andm 15=ωA2-42,∴A2-42A2-32=1520⇒16(A2-16)=9(A2-9)⇒A=5cm
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