If T is the time period of a simple pendulum, then at a time T/6 after the pendulum passes its mean position

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q92 – AP-EAMCET 2025

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Home Physics OscillationsQuestion #381

If T is the time period of a simple pendulum, then at a time

\frac{T}{6}

after the pendulum passes its mean position

Clue 1 of 3: Simple Harmonic Motion (SHM) Equations

Recall the equations for displacement, velocity, and acceleration of a particle undergoing SHM, especially when starting from the mean position.

🧠 Full Solution Path

Step 1: Identify the governing equations for SHM✦ Active

The motion of a simple pendulum for small oscillations is Simple Harmonic Motion (SHM). When the pendulum starts from its mean position (equilibrium position), its displacement $x$ at time $t$ is given by:

x = A \sin (ω t)

The velocity $v$ is the time derivative of displacement:

v = \frac{d x}{d t} = A ω \cos (ω t)

The acceleration $a$ is the time derivative of velocity:

a = \frac{d v}{d t} = - A ω^{2} \sin (ω t)

The maximum velocity is $v_{m a x} = A ω$ (occurs at $x = 0$ ). The maximum acceleration is $a_{m a x} = A ω^{2}$ (occurs at $x = \pm A$ ). The angular frequency $ω$ is related to the time period $T$ by $ω = \frac{2 π}{T}$ .

💡 Teacher's Secret Hint

Remember that the equations for SHM depend on the initial conditions. For starting from the mean position, use sine for displacement. If starting from an extreme position, use cosine.

Step 2: Calculate the phase angle at the given time○ Expand

The given time is $t = \frac{T}{6}$ . Substitute this into the expression for the phase angle $ω t$ :

ω t = (\frac{2 π}{T}) \times (\frac{T}{6}) = \frac{2 π}{6} = \frac{π}{3} radians

Step 3: Evaluate displacement, velocity, and acceleration at this phase angle○ Expand

Using the equations from Step 1 and the phase angle from Step 2:

1. **Displacement:**

x = A \sin (\frac{π}{3}) = A \frac{\sqrt{3}}{2}

2. **Velocity:**

v = A ω \cos (\frac{π}{3}) = A ω (\frac{1}{2})

Since $v_{m a x} = A ω$ , we have $v = \frac{1}{2} v_{m a x}$ .

3. **Acceleration:**

a = - A ω^{2} \sin (\frac{π}{3}) = - A ω^{2} \frac{\sqrt{3}}{2}

Since $a_{m a x} = A ω^{2}$ , we have $a = - \frac{\sqrt{3}}{2} a_{m a x}$ .

Step 4: Compare the results with the given options○ Expand

Let's check each option:

1. **Kinetic and potential energies of the pendulum are equal:**

K E = \frac{1}{2} m v^{2} = \frac{1}{2} m {(\frac{A ω}{2})}^{2} = \frac{1}{8} m A^{2} ω^{2}

P E = \frac{1}{2} m ω^{2} x^{2} = \frac{1}{2} m ω^{2} {(A \frac{\sqrt{3}}{2})}^{2} = \frac{1}{2} m ω^{2} A^{2} \frac{3}{4} = \frac{3}{8} m A^{2} ω^{2}

Since $K E \neq P E$ , option 1 is incorrect.

2. **The displacement of the pendulum is half of its amplitude ( $x = A / 2$ ):** We found $x = A \frac{\sqrt{3}}{2}$ . So, option 2 is incorrect.

3. **Acceleration of the pendulum is half of its maximum acceleration ( $a = a_{m a x} / 2$ ):** We found $a = - \frac{\sqrt{3}}{2} a_{m a x}$ . So, option 3 is incorrect.

4. **Velocity of the pendulum is half of its maximum velocity ( $v = v_{m a x} / 2$ ):** We found $v = \frac{1}{2} v_{m a x}$ . So, option 4 is correct.

💡 Teacher's Secret Hint

Pay attention to the signs for acceleration. While the magnitude is $\frac{\sqrt{3}}{2} a_{m a x}$ , the option only refers to 'half of its maximum acceleration' which implies magnitude. However, even by magnitude, it's not half.

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Physics Question 92 – AP-EAMCET 2025

Clue 1 of 3: Simple Harmonic Motion (SHM) Equations

🧠 Full Solution Path