Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 4 (Q.No. 4)
4.
A particle moves with simple harmonic motion. If its acceleration at distance 'D' from the equilibrium position is 'A' then the period of the motion is given by
2πAD
Discussion:
1 comments Page 1 of 1.
Bhupendra sahu said:
10 years ago
Answer is D.
T = 2pi/w.
w = Angular velocity.
Now, A = -w^2*d.
w = [A/d]^1/2.
T = 2pi[d/A]^1/2.
T = 2pi/w.
w = Angular velocity.
Now, A = -w^2*d.
w = [A/d]^1/2.
T = 2pi[d/A]^1/2.
(2)
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