Oscillations

JEE Main level, numerical answer, medium

19 of 194

Question 19 of 194

A particle executes simple harmonic motion with an amplitude of 4cm4cm. At the mean position, velocity of the particle is 10cms−110cm{s}^{-1}. The distance of the particle from the mean position when its speed becomes 5cms−15cm{s}^{-1} is αcm\sqrt{\alpha }cm, where α=\alpha =______.

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Answer: 12

The formula to calculate the velocity of the particle at the mean position is given by
V=Aω...(1)V=A\omega ...\left(1\right)
From equation (1), it follows that
10=4×ω⇒ω=52rads−110=4\times \omega \Rightarrow \omega =\dfrac{5}{2}rad{s}^{-1}
The formula to calculate the velocity of the particle at any instantaneous position xx can be expressed as
v=ωA2−x2...(2)v=\omega \sqrt{{A}^{2}-{x}^{2}}...\left(2\right)
From equation (2), it follows that
5=5242−x2⇒x2=16−4=12⇒x=12cm5=\dfrac{5}{2}\sqrt{{4}^{2}-{x}^{2}} \Rightarrow {x}^{2}=16-4 =12 \Rightarrow x=\sqrt{12}cm
Hence, α=12\alpha =12.

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