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BITSAT 2018
Exam
BITSAT
Year
2018
Subject
Physics
Chapter
Oscillations

BITSAT 2018 · Physics · Oscillations

If x, v and a denote the displacement, the velocity and the...

If x,vx, v and aa denote the displacement, the velocity and the acceleration of a particle executing SHMSHM of time period TT. Then, which of the following does not change with time?

Options

  1. AaTx\dfrac{aT}{x}
  2. BaT+2\pivaT + 2\piv
  3. CaTv\dfrac{aT}{v}
  4. Da2T2+4π2v2a^{2}T^{2} + 4{\pi}^{2}v^{2}

Correct answer

A. aTx\dfrac{aT}{x}

Step-by-step solution

We know, instantaneous displacement,

x=rsin\omegatx = rsin\omegat

∵∵ Instantaneous velocity, v=dxdt=rωcos\omegatv = \dfrac{dx}{dt} = r\omega cos\omegat

So. aTx=−ω2x×Tx=−ω2T=−4π2T2×T\dfrac{aT}{x} = \dfrac{- {ω}^{2}x \times T}{x} = - {ω}^{2}T = \dfrac{- 4{\pi}^{2}}{T2} \times T

=−4π2T== \dfrac{- 4{\pi}^{2}}{T} = constant

and,

⇒aTV=−ω2rsin\omegat×Tωrcos\omegat=−ωTtan\omegaT⇒ \dfrac{aT}{V} = \dfrac{- {ω}^{2}rsin\omegat \times T}{\omega rcos\omegat} = - \omega Ttan\omegaT

=−2πT×Ttan\omegat== - \dfrac{2\pi}{T} \times Ttan\omegat = not constant

Similarly, aT+2\pivaT + 2\piv and a2T2+4π2v2a^{2}T^{2} + 4{\pi}^{2}v^{2} is also not constant,
i.e. both are function of tt

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