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A particle executes simple harmonic motion with an amplitude of 4cm. At mean position the velocity of the particle is 10cm/sec. The distance of the particle from the mean position when its speed becomes 5 cm/s is

A particle executes simple harmonic motion with an amplitude of 4cm. At mean position the velocity of the particle is 10cm/sec. The distance of the particle from the mean position when its speed becomes 5 cm/s is

Grade:11

1 Answers

Raghav Jha
37 Points
6 years ago
As we know in SHM100=16 Omega,henceOmega=100\16(from given value)Now,Again ,5=100/16√A(square)-X(square)Hence X=3.37cm from mean position

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