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Question
A particle performing linear S.H.M. has the maximum velocity of 25 cm/s and maximum acceleration of 100 cm/ m2. Find the amplitude and period of oscillation. (π = 3.142)
Solution
In S.H.M the velocity is given as
Velocity, `v = A w cos wt,` where w is angular frequency and A is amplitude
Maximum velocity `vm = A w [therefore cos wt = 1]`
Acceleration `= -Aw^2 sin wt`
Maximum acceleration `a_m = |-A w^2| = Aw^2 (therefore sinw^2t = 1)`
`a_m = 100 cm//s^2, v_m = 25 cm//s`
`Aw = 25` ....(i)
`Aw^2 = 100` .....(ii)
Dividing (ii) by (i) `(Aw^2)/(Aw) = 100/25`
w = 4
Time period, T = `(2≠)/w`
`= (2xx3.14)/4`
`T = 1.57 sec`
from 1 wa = 25
a = `25/4 = 6.25 `cm
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