An object of mass 0.2kg executes SHM along the x-axis with a frequency of 25/π Hz. At the position x=0.04m the object has kinetic energy 0.5J and potential energy 0.4J. The amplitude of oscillation will be

• (A) 0.06m
• (B) 0.04m
• (C) 0.05m
• (D) 0.25m

To find the amplitude of oscillation for the object executing simple harmonic motion (SHM), we can use the relationship between kinetic energy (KE), potential energy (PE), and total mechanical energy (E) in SHM.

Key Formulas

The total mechanical energy in SHM is given by:

• E = KE + PE

In this case:

• KE = 0.5 J
• PE = 0.4 J

Calculating Total Energy

Now, we can calculate the total energy:

• E = 0.5 J + 0.4 J = 0.9 J

Potential Energy in SHM

The potential energy at a position x in SHM is given by:

• PE = (1/2) k x²

Where k is the spring constant. We can also express k in terms of the amplitude A and the total energy E:

• E = (1/2) k A²

Finding the Spring Constant

From the potential energy formula, we can rearrange it to find k:

• 0.4 J = (1/2) k (0.04 m)²

Solving for k:

• 0.4 = (1/2) k (0.0016)
• k = 500 N/m

Relating Total Energy to Amplitude

Now, substituting k back into the total energy equation:

• 0.9 J = (1/2) (500) A²

Solving for A:

• 0.9 = 250 A²
• A² = 0.0036
• A = 0.06 m

Final Answer

The amplitude of oscillation is 0.06 m, which corresponds to option (A).