A particle is moving in a straight line with SHM. Its velocity has the values $3 m / s$ and $2 m / s$ when its distances from mean position are $1 m$ and $2 m$ respectively. Find the length of its path and period of its motion.

$2.53 m$ , $4.87 s$

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$5.06 m$ , $6.03 s$

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$5.06 m$ , $4.87 s$

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None of these

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Solution

The correct option is C
$5.06 m$ , $4.87 s$

$v_{1} = 3 m / s a t x_{1} = 1 m a n d v_{2} = 2 m / s a t x_{2} = 2 m u s i n g v^{2} = ω^{2} (A^{2} - x^{2})$

We have:
$9 = ω^{2} (A^{2} - 1)$

$4 = ω^{2} (A^{2} - 4)$ on solving these

We get:

$A = \sqrt{6.4} = 2.53, a n d ω = \sqrt{\frac{5}{3}} = 1.29 r a d / s e c$

Hence, length of path $= 2 A = 5.06 m T i m e p e r i o d = T = \frac{2 π}{ω} = 4.87 s$

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A particle is moving in a straight line with SHM. Its velocity has the values 3 m/s and 2 m/s when its distances from mean position are 1 m and 2 m respectively. Find the length of its path and period of its motion.

The correct option is C 5.06 m, 4.87 s v1=3 m/s at x1=1 m and v2=2 m/s at x2=2 musing v2=ω2(A2−x2) We have: 9=ω2(A2−1) 4=ω2(A2−4) on solving these We get: A=√6.4=2.53, and ω=√53=1.29 rad/sec Hence, length of path=2A=5.06 mTime period=T=2πω=4.87 s

A particle is moving in a straight line with SHM. Its velocity has the values $3 m / s$ and $2 m / s$ when its distances from mean position are $1 m$ and $2 m$ respectively. Find the length of its path and period of its motion.