The total energy of a particle executing S.H.M. is 80 J . What is the potential energy when the particle is at a distance of 3/4 of amplitude from the mean position

Options

A. 60 j
B. 10 j
C. 40 j
D. 45 j

Answer

To solve the problem, we need to determine the potential energy of a particle executing Simple Harmonic Motion (SHM) when it is at a distance of \( \frac{3}{4} \) of the amplitude from the mean position.

### Step-by-Step Solution:

1. **Understand the Total Energy in SHM**:
The total mechanical energy \( E \) of a particle in SHM is constant and is given by the sum of its kinetic energy \( KE \) and potential energy \( PE \).
\[ E = KE + PE \]
Given that the total energy \( E = 80 \, J \).

2. **Identify the Position**:
The particle is at a distance of \( x = \frac{3}{4}A \) from the mean position, where \( A \) is the amplitude of the motion.

3. **Calculate Potential Energy**:
The potential energy \( PE \) at a distance \( x \) in SHM is given by the formula:
\[ PE = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant. However, we can also express \( PE \) in terms of total energy and amplitude:
\[ PE = E - KE \]
The kinetic energy \( KE \) can be expressed as:
\[ KE = \frac{1}{2} k (A^2 - x^2) \]
Since \( x = \frac{3}{4}A \), we can substitute:
\[ KE = \frac{1}{2} k \left(A^2 - \left(\frac{3}{4}A\right)^2\right) \]
Simplifying this:
\[ KE = \frac{1}{2} k \left(A^2 - \frac{9}{16}A^2\right) = \frac{1}{2} k \left(\frac{16}{16}A^2 - \frac{9}{16}A^2\right) = \frac{1}{2} k \left(\frac{7}{16}A^2\right) \]

4. **Relate \( k \) to Total Energy**:
The total energy can also be expressed in terms of amplitude:
\[ E = \frac{1}{2} k A^2 \]
Setting this equal to the total energy given:
\[ 80 = \frac{1}{2} k A^2 \implies k A^2 = 160 \]

5. **Substituting \( k \) into KE**:
Now we substitute \( k \) back into the equation for \( KE \):
\[ KE = \frac{1}{2} \left(\frac{160}{A^2}\right) \left(\frac{7}{16}A^2\right) = \frac{160 \cdot 7}{32} = 35 \, J \]

6. **Calculate Potential Energy**:
Now, we can find the potential energy using the total energy:
\[ PE = E - KE = 80 - 35 = 45 \, J \]

### Final Answer:
The potential energy when the particle is at a distance of \( \frac{3}{4} \) of the amplitude from the mean position is \( 45 \, J \).