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A particular executes SHM with a time pe...

A particular executes SHM with a time period of 16s. At time `t=2s`, the particle crosses the mean position while at t= 4 , its velocity is `4ms^(-1)` Find its a amplitude of motion.

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Here,
`T=25s, "At "t=2s,y=0 and "at "t =5s v = 4 ms^(-1), a = ? `
For simple harmonic motion , y = a sin wt = a sin `(2pi)/(T) t` when t = 4 s , the time taken by particle to travel from the mean position to a given position
`y=asin ((2pi)/(16)xx2)=a sin ((x)/(4)) =a/(sqrt(2))`
Velocity `v=wsqrt(a^(2)-y^(2))`
`4=((2pi)/(16))sqrt(a^(2)-(a^(2))/(2))`
`=(pi)/(8)xx(a)xxsqrt(2)`
`a=(32xxsqrt(2))/(pi)=14.44`
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