First slide

Simple hormonic motion

Question

A particle is oscillating along a straight line with time period 6 sec and amplitude ‘A’. Then the time taken by the particle to directly move from $x = \frac{- A}{\sqrt{2}}$ to $x = + \frac{A}{2}$ is

Difficult

A

2 sec
B

1 sec
C

1.5 sec
D

1.25 sec

Solution

In the phasor diagram,
$OP = \frac{A}{\sqrt{2}} and OQ = \frac{A}{2}$
$∴ ∠ MOP = \cos^{- 1} (\frac{OP}{OM}) = \cos^{- 1} \frac{1}{\sqrt{2}} = \frac{π}{4} and ∠ NOP = \cos^{- 1} (\frac{OQ}{ON}) = \cos^{- 1} (\frac{1}{2}) = \frac{π}{3}$
$∴ θ = π - (\frac{π}{4} + \frac{π}{3}) = \frac{5 π}{12}$
$∴$ Time taken by the reference particle to move from M to N $= \frac{θ}{ω} = \frac{5 π / 12}{2 π / T} = \frac{5 T}{24} = \frac{5 \times 6}{24} = 1 .25 \sec$ .

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