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Physics - SHM Question with Solution | TestHub

PhysicsSHMEnergyMedium3 minQB

PhysicsMediummultiple choice

The energy of a particle executing simple harmonic motion is given by $E = Ax^{2} + Bv^{2}$ where x is the displacement from mean position $x = 0$ and v is the velocity of the particle at x then choose the correct statement(s).

Options:(select one or more)

Answer:

B, C

Solution:

Compare $E = A x^{2} + B v^{2}$ with $E = \frac{1}{2} k x^{2} + \frac{1}{2} m v^{2}$ : $k = 2 A$ , $m = 2 B$ .

At the extreme $E = A x_{m a x}^{2}$ , so amplitude $E / A$ , not $2 E / A$ . (A) false.

$ω = k / m = A / B$ , $T = 2 π B / A$ . These match (B) and (C).

Answer: B,C

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Energy

⏱ 3mℹ️ Source: QB

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