Question

The energy of a particle executing Simple Harmonic Motion (SHM) is given by E = Ax^2 + BV^2. Here 'x' is the displacement of the particle from its mean position, 'V' is its velocity at 'x', and A and B are positive constants. The maximum velocity of the particle is:

• A. \( \sqrt{\frac{E}{B}} \)
• B. \( \sqrt{\frac{E}{A}} \)
• C. \( \sqrt{\frac{2E}{B}} \)
• D. \( \sqrt{\frac{2E}{A}} \)

Hint:

In SHM: \[ v_{\max} \text{ occurs at mean position where } x = 0 \]

Answer 1

The Correct Option is A

Step 1: Read the energy expression.
The total energy is given as $E = Ax^2 + Bv^2$. Here $x$ is the displacement and $v$ is the speed at that point, while $A$ and $B$ are fixed positive numbers.

Step 2: Recognise the two parts.
In a vibrating system, total energy is potential energy plus kinetic energy. The part with $x^2$ acts like potential energy, and the part with $v^2$ acts like kinetic energy.

Step 3: Find where speed is largest.
The speed is biggest at the mean position, where the displacement is zero. So set $x = 0$.

Step 4: Simplify the energy there.
With $x = 0$, the first term vanishes and all the energy is in the speed term: \[ E = B\,v_{\max}^2. \]

Step 5: Solve for the maximum speed.
Divide by $B$ and take the square root: \[ v_{\max}^2 = \frac{E}{B}, \qquad v_{\max} = \sqrt{\frac{E}{B}}. \]

Step 6: State the answer.
The largest speed of the particle is the square root of $E$ over $B$. \[ \boxed{\sqrt{\dfrac{E}{B}}} \]