A particle executing SHM along a straight line has zero velocity at points (A ) and (B ) whose distance from (O ) on the same line (O A B ) are (a ) and (b ), respectively. If the velocity at the mid point between (A ) and (B ) is (v ), the

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A particle executing SHM along a straight line has zero velocity at points (A ) and (B ) whose distance from (O ) on the same line (O A B ) are (a ) and (b ), respectively. If the velocity at the mid point between (A ) and (B ) is (v ), then its time period is

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Correct answer: B. ( ( b-a v ) )

According to the question, ( ) Amplitude (= Distance travelled by the particles 2 (A to B) ) Amplitude of particles executing simple harmonic motion (SHM) along a straight line (A B ) is ((a)= b-a 2 ). Velocity of particle, (v= ) Amplitude ( ) Oscillation frequency ( aligned v & =a v & = ( b-a 2 ) or = 2 v b-a (i) aligned ) ( ) Time period, (T= 2 ) Putting the value of ( ) from Eq. (i) to above formula, ( T= 2 2 v (b-a) T= b-a v ) So, the time period of a particle executing SHM along a straight line from points (A

A particle executing SHM along a straight line has zero velocity at points (A ) and (B ) whose distance from (O ) on the same line (O A B ) are (a ) and (b ), respectively. If the velocity at the mid point between (A ) and (B ) is (v ), then its time period is

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