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A particle performs SHM in a straight li...

A particle performs SHM in a straight line. In the first second, starting from rest, it travels a distance a and in the next second it travels a distance b in the same side of mean position. The amplitude of the SHM is

A

a-b

B

`(2a-b)/(3)`

C

`(2a^(2))/(3a-b)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
(c) Particle starts from rest. Hence, `x=A "cos" omegat`
`a=A-A"cos"(omegaxx1)`
or ` "cos"omega=(A-a)/(A)=(1-(a)/(A))`
`implies a+b=A-A"cos"(omegaxx2)`
`=A-A(2"cos"^(2)omega-1)`
`=2A=2A(1-(a)/(A))^(2)`
ON Solving, we get `A=(2a^(2))/(3a-b)`
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