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Question
A particle is performing S.H.M. of amplitude 5 cm and period of 2s. Find the speed of the particle at a point where its acceleration is half of its maximum value.
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Solution
Given:
A = 5 cm = 5 × 10−2 m, T = 2 s,
a = `"a"_"max"/2 = ("Aomega^2)/2` ….(i)
To find: speed (v)
Formula: v = `omegasqrt("A"^2 - "x"^2)`
Calculation:
Since, a = ω2x
∴ x = `"a"/ω^2 = ("A"ω^2)/(2ω^2) = "A"/2` ….[from (i)]
From formula,
v = `ωsqrt("A"^2 - "A"^2/4) = sqrt3/2"A"ω`
= `sqrt3/2 xx 5 xx 10^-2 xx (2pi)/"T"` .....`[∵ ω = (2pi)/"T"]`
= `sqrt3/2 xx 5 xx 10^-2 xx (2 xx 3.14)/2`
∴ v = 13.6 × 10−2 m/s
The speed of the particle where its acceleration is half of its maximum value is 13.6 × 10−2 m/s.
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