A particle describes linear S.H.M. Its velocities are $3 m s^{- 1}$ and $2 m s^{- 1}$ when its displacements are 2 m and 3 m respectively from mean position. Find length of the path?

1.2 m

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4.6 m

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6.7 m

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7.2 m

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Solution

The correct option is D 7.2 m
$v = ω \sqrt{A^{2} - x^{2}}; \frac{3}{2} = \frac{\sqrt{A^{2} - 4}}{\sqrt{A^{2} - 9}}$
Squaring, $\frac{9}{4} = \frac{(A^{2} - 4)}{(A^{2} - 9)}$
$9 A^{2} - 81 = 4 A^{2} - 16; 5 A^{2} = 65$
$A^{2} = \frac{65}{5} = 13; A = \sqrt{13} m$
$∴$ Length of the path $= 2 A = 2 \sqrt{13} = 7.2 m .$

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A particle describes linear S.H.M. Its velocities are 3ms−1 and 2ms−1 when its displacements are 2 m and 3 m respectively from mean position. Find length of the path?

The correct option is D 7.2 mv=ω√A2−x2;32=√A2−4√A2−9 Squaring, 94=(A2−4)(A2−9) 9A2−81=4A2−16;5A2=65 A2=655=13;A=√13m ∴ Length of the path =2A=2√13=7.2m.

A particle describes linear S.H.M. Its velocities are $3 m s^{- 1}$ and $2 m s^{- 1}$ when its displacements are 2 m and 3 m respectively from mean position. Find length of the path?