Question

The energy of a particle executing SHM is given by E = Ax^2 + BV^2. The INCORRECT statement is:

• A. Amplitude is EA
• B. Maximum velocity is EB
• C. Time period is 2BA
• D. Maximum acceleration is EAB

Hint:

Always evaluate SHM energy questions using two extreme cases: - x = A (velocity zero) - x = 0 (velocity maximum)

Answer 1

The Correct Option is A

Step 1: Read the energy expression.
The total energy in SHM is given as $E = A x^2 + B v^2$, where $x$ is the displacement and $v$ is the velocity. We must find the statement that is NOT correct.

Step 2: Compare with the standard energy form.
The usual SHM energy is the sum of potential and kinetic parts: \[ E = \frac{1}{2} m \omega^2 x^2 + \frac{1}{2} m v^2 \] Matching terms gives $A = \frac{1}{2} m \omega^2$ and $B = \frac{1}{2} m$.

Step 3: Check the amplitude claim.
At the extreme point the velocity is zero and displacement equals the amplitude $R$. So $E = A R^2$, which gives \[ R = \sqrt{\frac{E}{A}} \] The amplitude is a square root, not simply $\frac{E}{A}$ as the first option states. So that option is wrong.

Step 4: Check the maximum velocity claim.
At the mean point displacement is zero and velocity is largest. So $E = B v_{max}^2$, giving $v_{max} = \sqrt{\frac{E}{B}}$. This is a genuine result, so that statement is acceptable.

Step 5: Check time period and acceleration.
Using $A$ and $B$, the ratio $\frac{A}{B} = \omega^2$, so the time period and the maximum acceleration come out in correct forms involving these constants. These statements describe real SHM behaviour.

Step 6: Pick the incorrect statement.
Only the amplitude statement fails, because the true amplitude is $\sqrt{E/A}$ and not $E/A$. So the incorrect statement is the one that claims the amplitude is $\frac{E}{A}$. \[ \boxed{\text{Amplitude is } \tfrac{E}{A}} \]