A particle is performing SHM energy of vibration 90J and amplitude 6cm. When the particle reaches at distance 4cm from mean position, it is stopped for a moment and then released. The new energy of vibration will be

Options

A. 40J
B. 50J
C. 90J
D. 60J

Answer

To solve the problem step by step, we will analyze the situation of a particle performing Simple Harmonic Motion (SHM) and how its energy changes when it is stopped and then released from a certain position.

### Step 1: Understand the Energy in SHM
The total mechanical energy (E) of a particle in SHM is given by the formula:
\[ E = \frac{1}{2} k A^2 \]
where:
- \( k \) is the spring constant,
- \( A \) is the amplitude of the motion.

Given that the total energy \( E = 90 \, J \) and the amplitude \( A = 6 \, cm = 0.06 \, m \).

### Step 2: Calculate the Spring Constant (k)
We can rearrange the energy formula to find \( k \):
\[ k = \frac{2E}{A^2} \]
Substituting the known values:
\[ k = \frac{2 \times 90}{(0.06)^2} \]
\[ k = \frac{180}{0.0036} \]
\[ k = 50000 \, N/m \]

### Step 3: Determine the New Amplitude After Stopping
When the particle is at a distance of \( 4 \, cm = 0.04 \, m \) from the mean position and is stopped, it will have potential energy at that position. The potential energy (PE) at that position is given by:
\[ PE = \frac{1}{2} k x^2 \]
where \( x = 0.04 \, m \).

Calculating the potential energy:
\[ PE = \frac{1}{2} \times 50000 \times (0.04)^2 \]
\[ PE = \frac{1}{2} \times 50000 \times 0.0016 \]
\[ PE = 40 \, J \]

### Step 4: Determine the New Energy of Vibration
When the particle is released from this position, all the potential energy will convert back to total mechanical energy. Since the particle was stopped, the total energy when released will now be equal to the potential energy calculated at \( 4 \, cm \):
\[ E_{new} = PE = 40 \, J \]

### Conclusion
The new energy of vibration when the particle is released from \( 4 \, cm \) will be:
\[ \text{New Energy} = 40 \, J \]

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