A particle is executing S.H.M. of amplitude 5cm 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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Answer:
13.6 cm/ sec
Explanation:
a=-w²x
a(max)= -w²A
= -w²5
We want accⁿ to be half of its maximum value
a = (-w²5)/2
a = -w²5/2
Comparing with -w²x to get the point at which its accⁿ is half the original value
-w²x = -w²5/2
x = 5/2
Putting value of x in formula
v = w√(A²-x²)
v = w√(5²-(5/2)²)
=w√(25-6.25)
v = w√18.75 ———(i)
Calculating value of w by time period
T = 2π/w
w = 2π/T
w = 2π/2 = π
Putting value of w in eq. i
v = π√18.75 ≈ 13.6 cm/sec
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