Given: Total energy of a particle executing simple harmonic motion = 80 J, distance from mean position = 3/4 of amplitude.

To find: Potential energy of the particle at the given position.

Solution:

1. Formula for total energy in SHM:

Total energy of a particle executing SHM is given by the formula:

E = Kinetic energy + Potential energy

where,

Kinetic energy = (1/2)mv² (m is the mass of the particle and v is its velocity)

Potential energy = (1/2)kx² (k is the spring constant and x is the displacement of the particle from its mean position)

2. Kinetic energy in SHM:

At any point in SHM, the velocity of the particle is given by:

v = ±ω√(A² - x²)

where,

ω is the angular frequency of SHM

A is the amplitude of SHM

x is the displacement of the particle from its mean position

Therefore, the kinetic energy of the particle is given by:

K.E. = (1/2)mv² = (1/2) m ω² (A² - x²)

3. Potential energy in SHM:

The potential energy of the particle is given by:

P.E. = (1/2)kx²

4. Finding potential energy at the given position:

Given that the total energy of the particle is 80 J.

Therefore, at any position x,

E = K.E. + P.E.

80 = (1/2) m ω² (A² - x²) + (1/2)kx²

Since the mass of the particle and angular frequency are constant, we can write:

80 = constant + (1/2)kx²

At x = 3/4 A, the displacement of the particle is:

x = (3/4)A

Substituting this value in the above equation, we get:

80 = constant + (1/2)k(3/4A)²

Simplifying,

80 = constant + (9/32)kA²

Since the constant is unknown, we cannot find the value of k. However, we can find the value of potential energy at the given position using the formula:

P.E. = (1/2)kx²

At x = 3/4 A, we have:

P.E. = (1/2)k(3/4A)² = (9/32)kA²

Substituting the value of (9/32)kA² in the above equation, we get:

P.E. = 45 J

Therefore, the potential energy of the particle at a distance of 3/4 of amplitude from the mean position is 45 J.

Hence, the correct option is (d) 45 J.