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 follow these steps:

### Step 1: Understand the Given Information
We are given:
- Total energy (E) of the particle performing SHM = 90 J
- Amplitude (A) = 6 cm
- Distance from the mean position when stopped (x) = 4 cm

### Step 2: Determine the New Amplitude
When the particle is stopped at a distance of 4 cm from the mean position, the new amplitude (A') will be equal to this distance, which is:
\[ A' = 4 \, \text{cm} \]

### Step 3: Use the Formula for Energy in SHM
The total energy in SHM is given by the formula:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
Where:
- E = total energy
- m = mass of the particle
- ω = angular frequency
- A = amplitude

### Step 4: Relate the New Energy to the Original Energy
The new energy (E') when the amplitude is A' can be expressed as:
\[ E' = \frac{1}{2} m \omega^2 (A')^2 \]

### Step 5: Set Up the Ratio of Energies
We can set up the ratio of the new energy to the original energy:
\[ \frac{E'}{E} = \frac{(A')^2}{A^2} \]

### Step 6: Substitute the Values
Substituting the values we have:
- \( A' = 4 \, \text{cm} \)
- \( A = 6 \, \text{cm} \)
- \( E = 90 \, \text{J} \)

So,
\[ \frac{E'}{90} = \frac{(4)^2}{(6)^2} \]
\[ \frac{E'}{90} = \frac{16}{36} \]

### Step 7: Solve for the New Energy
Now, we can solve for \( E' \):
\[ E' = 90 \times \frac{16}{36} \]
\[ E' = 90 \times \frac{4}{9} \]
\[ E' = 40 \, \text{J} \]

### Final Answer
The new energy of vibration will be **40 J**.

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