Time Period of Pendulum — Revision Notes

⚡ 30-Second Revision

Formula: — $T = 2pisqrt{\frac{L}{g}}$
T depends on: — Length ($L$) and acceleration due to gravity ($g$).
T is independent of: — Mass of bob ($m$) and amplitude (for small angles, $heta < 15^circ$).
Proportionalities: — $T propto sqrt{L}$, $T propto \frac{1}{sqrt{g}}$.
In a lift (upward acc. $a$): — $g_{eff} = g+a implies T$ decreases.
In a lift (downward acc. $a$): — $g_{eff} = g-a implies T$ increases.
Free fall: — $g_{eff} = 0 implies T = infty$ (no oscillation).
In a fluid: — $g_{eff} = g(1 - \frac{\rho_f}{\rho_b}) implies T$ increases.
Temperature effect: — $Delta L = LalphaDelta\theta implies T' approx T(1 + \frac{1}{2}alphaDelta\theta)$. Increase in temp $implies$ increase in $L implies$ increase in $T implies$ clock loses time.

2-Minute Revision

The time period of a simple pendulum, $T$, is the time for one complete oscillation. The fundamental formula is $T = 2pisqrt{L/g}$, where $L$ is the effective length and $g$ is the acceleration due to gravity.

Crucially, for small angles (typically less than $15^circ$), the time period is independent of the bob's mass and the amplitude of oscillation. This is a common trap in NEET questions. The time period increases with increasing length ($T propto sqrt{L}$) and decreases with increasing gravity ($T propto 1/sqrt{g}$).

Remember how $g$ changes in different scenarios: in an upward accelerating lift, $g_{eff} = g+a$, making $T$ shorter; in a downward accelerating lift, $g_{eff} = g-a$, making $T$ longer. In free fall, $g_{eff}=0$, so the pendulum doesn't oscillate ($T=infty$).

If the pendulum is in a fluid, buoyancy reduces the effective gravity, increasing $T$. Thermal expansion of the string due to temperature changes can also alter $L$, thus affecting $T$ and the clock's accuracy.

5-Minute Revision

A simple pendulum's time period, $T$, is a measure of how long it takes to complete one full swing. The bedrock of its understanding lies in the formula $T = 2pisqrt{L/g}$. Here, $L$ is the length from the pivot to the bob's center of mass, and $g$ is the local acceleration due to gravity. This formula is valid under the 'small angle approximation' (typically $heta < 15^circ$), which ensures the motion is Simple Harmonic Motion (SHM).

Key takeaways for NEET:

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Dependencies: — $T$ is directly proportional to the square root of $L$ ($T propto sqrt{L}$) and inversely proportional to the square root of $g$ ($T propto 1/sqrt{g}$). This means a longer pendulum swings slower, and stronger gravity makes it swing faster.
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Independencies: — For small oscillations, $T$ is independent of the bob's mass and the amplitude of oscillation. This is a frequent conceptual question.
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**Variations in $g$:**

* In a lift: If a lift accelerates upwards with $a$, $g_{eff} = g+a$, so $T$ decreases. If it accelerates downwards with $a$, $g_{eff} = g-a$, so $T$ increases. In free fall ($a=g$), $g_{eff}=0$, and $T$ becomes infinite (no oscillation).

* Altitude/Depth: At higher altitudes, $g$ decreases, so $T$ increases. At the center of Earth, $g=0$, so $T=infty$. * In a fluid: The buoyant force reduces the effective weight, leading to $g_{eff} = g(1 - \frac{\rho_f}{\rho_b})$, where $ho_f$ is fluid density and $ho_b$ is bob density.

Since $g_{eff} < g$, $T$ increases.

1
Thermal Expansion: — If temperature changes, the length $L$ of the pendulum rod changes ($Delta L = LalphaDelta\theta$). This affects $T$. An increase in temperature increases $L$, which increases $T$, causing a clock to lose time. The fractional change in period is $rac{Delta T}{T} approx \frac{1}{2}alphaDelta\theta$.

Mini-Example: If a pendulum clock runs slow by $10,\text{s}$ in a day, it means its time period has increased. To correct it, you need to decrease its time period, which means you need to shorten its length slightly.

Prelims Revision Notes

The time period of a simple pendulum is a critical topic for NEET, primarily tested through its formula and dependencies. The fundamental formula is $T = 2pisqrt{\frac{L}{g}}$, where $L$ is the effective length (pivot to center of mass of bob) and $g$ is the local acceleration due to gravity. This formula holds for small angular displacements, typically less than $10^circ$ to $15^circ$, where the motion approximates Simple Harmonic Motion (SHM).

Key Relationships:

$T propto sqrt{L}$: Longer pendulums have longer periods (swing slower).
$T propto \frac{1}{sqrt{g}}$: Stronger gravity leads to shorter periods (swings faster).

Crucial Independencies (Common Traps):

Mass of the bob: — The time period is independent of the mass of the bob. This is because both the restoring force and inertia are proportional to mass, causing it to cancel out in the equation of motion.
Amplitude of oscillation: — For small angles, the time period is independent of the amplitude. For larger amplitudes, $T$ actually increases, and the motion is no longer perfectly SHM.

Scenarios Affecting Time Period:

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In an accelerating lift:

* Lift accelerating upwards with $a$: $g_{eff} = g+a$. $T$ decreases. * Lift accelerating downwards with $a$: $g_{eff} = g-a$. $T$ increases. * Lift in free fall ($a=g$): $g_{eff} = 0$. $T \to infty$ (no oscillation).

1
Effect of Altitude/Depth:

* Higher altitude: $g$ decreases, so $T$ increases. * At Earth's center: $g=0$, so $T \to infty$.

1
In a fluid medium: — The buoyant force reduces the effective weight. $g_{eff} = g(1 - \frac{\rho_f}{\rho_b})$, where $ho_f$ is fluid density and $ho_b$ is bob density. Since $g_{eff} < g$, $T$ increases.
2
Temperature Change (Thermal Expansion): — If the temperature increases, the length $L$ of the pendulum rod increases ($Delta L = LalphaDelta\theta$). This leads to an increase in $T$. The fractional change in period is $rac{Delta T}{T} = \frac{1}{2}alphaDelta\theta$. An increase in $T$ means the clock runs slower and loses time.

Seconds Pendulum: A pendulum with a time period of $2,\text{s}$. Its length can be calculated by setting $T=2$ in the formula.

Problem-Solving Tip: For problems involving changes in $L$ or $g$, use ratios: $rac{T_2}{T_1} = sqrt{\frac{L_2}{L_1}}$ or $rac{T_2}{T_1} = sqrt{\frac{g_1}{g_2}}$. This simplifies calculations significantly.

Vyyuha Quick Recall

Long Gravity, Short Time. Short Length, Short Time. Mass And Amplitude Don't Matter (for small angles).

(L = Length, G = Gravity, T = Time Period, M = Mass, A = Amplitude)