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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Physics
Chapter
Oscillations

BITSAT 2023 · Paper 2 · Physics · Oscillations

The variation of kinetic energy (KE) of a particle executing simple...

The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x) starting from mean position to extreme position (A) is given byThe variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x) startingThe variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x) startingThe variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x) startingThe variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x) starting

Options

  1. ATwo stones of masses m and 2m are whirled in horizontal circles, the heavier one in radius (r)/(2) and the lighter one
  2. BTwo stones of masses m and 2m are whirled in horizontal circles, the heavier one in radius (r)/(2) and the lighter one
  3. CTwo stones of masses m and 2m are whirled in horizontal circles, the heavier one in radius (r)/(2) and the lighter one
  4. DTwo stones of masses m and 2m are whirled in horizontal circles, the heavier one in radius (r)/(2) and the lighter one

Correct answer

D. Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in radius (r)/(2) and the lighter one

Step-by-step solution


T⋅E=K⋅E+P.E⇒K⋅E=T⋅E−K⋅E=12k2−12k2⇒KE=12mω2(A2−x2)[∵k=mω2]\begin{aligned} & T \cdot E=K \cdot E+P. E \\ & \Rightarrow K \cdot E=T \cdot E-K \cdot E \\ & =\frac{1}{2} k^2-\frac{1}{2} k^2 \\ & \Rightarrow K E=\frac{1}{2} m \omega^2\left(A^2-x^2\right) \quad\left[\because k=m \omega^2\right]\end{aligned}
KE\mathrm{KE} is maximum at x=0 KE\mathrm{x}=0 \ \mathrm{KE} is zero when x=A∴K.E\mathrm{x}=\mathrm{A}\therefore \mathrm{K}. \mathrm{E} vs x\mathrm{x} graph is parabola.
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