Question

The displacement of an object attached to a spring and excutes simple harmonic motion is given by x = 2×10⁻²cos2πt metre. The time at which the maximum speed first occures is ? ​

Answer 1

Answer:

0.25sec

Explanation:

Given :- x = 2× 10⁻²cos2πt

velocity 'v' = dx/dt = d(2×10⁻²cos2πt) /dt

= 2 × 2π ×10⁻² × sin2πt

so, the velocity will be maximum when sin2πt =1 ∵ Rε [0,1]

∴ sin2πt = sinπ/2 [for first maximum speed take 1 = sinπ/2]

• ∵ 2πt = π/2

• ⟹ 2t = 1/2

• ⟹ t = 1/4 = 0.25s Answer

The time at which the maximum speed first occurs is 0.25s

Answer 2

Given, displacement x=2×10^−2 cosπt

The magnitude of velocity , v=∣dx/dt∣

=2π×10^−2 sinπt

So, the velocity will be maximum when sinπt=1 where πt= π/2, 3π/2,..

So for first maximum, πt= π/2 or t=0.5s

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