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A simple pendulum executing SHM in a str...

A simple pendulum executing `SHM` in a straight line has zero velocity at 'A' and 'B' whose distances form 'O' in the same line `OAB` are 'a' and 'b'. If the velocity half wat between them is 'v' then its time period is

A

` ( 2pi ( b - a ))/( v ) `

B

` ( pi ( b - a )) / (v) `

C

` ( pi ( b + a )) /( v ) `

D

None

Text Solution

Verified by Experts

The correct Answer is:
B

Amp. ` = ( b - a ) / ( 2 ) `
` V_ (mp ) = Aomega rArr v = (( b - a ) ) /( 2 ) ( 2pi ) /( T) rArr T = ( pi ( b - a )) /( v ) `
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Knowledge Check

  • A particle is executing SHM along a straight line. Its velocities at distance x_(1)" and "x_(2) form the mean position are v_(1)" and "v_(2) , respectively. Its time period is………..

    A
    `2pi sqrt((x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)+v_(2)^(2)))`
    B
    `2pi sqrt((-x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)-v_(2)^(2)))`
    C
    `2pi sqrt((v_(1)^(2)+v_(2)^(2))/(x_(1)^(2)+x_(2)^(2)))`
    D
    `2pi sqrt((v_(1)^(2)-v_(2)^(2))/(x_(1)^(2)-x_(2)^(2)))`

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