Physics Q91 – AP-EAMCET 2026

A particle is executing simple harmonic motion with an amplitude of

25 cm

. If the kinetic energy of the particle at the mean position is

5 J

, then the maximum force acting on the particle in its motion is

Clue 1 of 3: Energy Conservation in SHM

In Simple Harmonic Motion, the total mechanical energy of the system remains constant. At the mean position, the potential energy is zero, and thus the total energy is equal to the maximum kinetic energy.

🧠 Full Solution Path

Step 1: Identify Given Parameters and Convert Units✦ Active

The problem provides the amplitude of simple harmonic motion and the kinetic energy at the mean position. We need to find the maximum force acting on the particle.

Given:

A = 25 cm = 0.25 m

K E_{m a x} = 5 J

💡 Teacher's Secret Hint

Always ensure all units are consistent (e.g., SI units) before performing calculations. Converting centimeters to meters is a crucial first step.

Step 2: Relate Maximum Kinetic Energy to SHM Parameters○ Expand

In Simple Harmonic Motion (SHM), the kinetic energy is maximum at the mean position. The formula for maximum kinetic energy is:

K E_{m a x} = \frac{1}{2} m ω^{2} A^{2}

Where $m$ is the mass of the particle, $ω$ is the angular frequency, and $A$ is the amplitude. We are given $K E_{m a x} = 5 J$ , so:

5 = \frac{1}{2} m ω^{2} A^{2} \Rightarrow m ω^{2} A^{2} = 10

💡 Teacher's Secret Hint

Remember that the total mechanical energy in SHM is $E = \frac{1}{2} k A^{2}$ , and $k = m ω^{2}$ . So, $E = \frac{1}{2} m ω^{2} A^{2}$ . At the mean position, $P E = 0$ , so $E = K E_{m a x}$ .

Step 3: Express Maximum Force in Terms of SHM Parameters○ Expand

The restoring force in SHM is given by $F = - k x$ , where $k = m ω^{2}$ . The magnitude of the force is $F = m ω^{2} x$ . The maximum force occurs at the extreme positions ( $x = \pm A$ ). Therefore, the maximum force ( $F_{m a x}$ ) is:

F_{m a x} = m ω^{2} A

💡 Teacher's Secret Hint

The maximum force is directly proportional to the amplitude and the square of the angular frequency, $F_{m a x} \propto A ω^{2}$ .

Step 4: Calculate the Maximum Force○ Expand

From Step 2, we have $m ω^{2} A^{2} = 10$ . We can rewrite this as $(m ω^{2} A) \cdot A = 10$ . Since $F_{m a x} = m ω^{2} A$ , we can substitute this into the equation:

F_{m a x} \cdot A = 10

Now, substitute the value of amplitude $A = 0.25 m$ :

F_{m a x} \cdot (0.25 m) = 10 J

Solve for $F_{m a x}$ :

F_{m a x} = \frac{10 J}{0.25 m} = \frac{10}{1 / 4} N = 10 \times 4 N = 40 N

💡 Teacher's Secret Hint

A quick way to check if your answer makes sense is through dimensional analysis. Energy is in Joules (N·m), and dividing by distance (m) gives Force (N).

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Physics Question 91 – AP-EAMCET 2026

Clue 1 of 3: Energy Conservation in SHM

🧠 Full Solution Path