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Question
The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the center of the path.
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Solution
Given:
m = 2 kg,
T.E. = 40 J
To find:
Speed while crossing the mean position (vmax) = ?
Formula:
T.E. = `1/2 mv_"max"^2`
Calculation:
From formula,
`v_"max" = sqrt((2 xx T.E.)/m)`
= `sqrt((2 xx 40)/2)`
= `2 sqrt 10`
= 2 × 3.162
= 6.324 m/s
The speed of the particle while crossing the mean position is 6.324 m/s.
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