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If the displacement `x` and velocity `v` of a particle executing `SHM` are related as `4v^(2) = 25 - x^(2)`. Then its maximum displacement in meter `(x, v` are in `SI`) is

A

1

B

2

C

5

D

6

Text Solution

Verified by Experts

The correct Answer is:
C
`v^(2) = (1)/(4) (25 - x^(2))`
`rArr v = (1)/(2) sqrt(5^(2) - x^(2))`
A = 5 m
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NARAYNA-OSCILLATIONS-EXERCISE - I (H.W)
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  11. The acceleration of a particle in SHM is 0.8ms^(-2), when its displace...

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  19. Due to some force F(1) a body oscillates with period 4//5s and due to ...

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