The amplitude of a particle executing SHM is 4 cm . At the mean position the speed of particle is 16 cm/s . The distance of the particle from the mean position at which the speed of particle becomes 8✓3 cm/s will be what ?

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Solution

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

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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