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Question: Answered & Verified by Expert
A point mass m=20 kg, is suspended by a massless spring of constant 2000 N/M. The point mass is released when elongation in the spring is 15 cm. The equation of displacement of particle as a function of time is (Take g=10 m/s2) 

PhysicsOscillationsNEET
Options:
  • A y=10sin10t
  • B y=10cos10t
  • C y=10sin10t+π6
  • D None of these
Solution:
1779 Upvotes Verified Answer
The correct answer is: y=10sin10t+π6
The motion of block is S.H.M.

y=Asinωt+ ϕ

Here amplitude is A=mgk

=20 ×102000m=10 cm

At t=0, displacement of body with respect to mean position is

y=15-10=5cm

5=10sinω×0+ ϕ

or 12=sinϕ    ϕ=π6

y=10sin10t+π6

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