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Q14.16(a):
Answer the following questions : (a) Time period of a particle in SHM depends on the force constant $k$ and mass $m$ of the particle: $T = 2\pi \sqrt{\frac{m}{k}}$. A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

Solution :

Step 1: Analyzing the Forces on a Simple Pendulum

Consider a simple pendulum consisting of a bob of mass $m$ suspended by a massless, inextensible string of length $l$. When the pendulum is displaced by an angle $\theta$ from the vertical, the restoring force $F$ is provided by the tangential component of the gravitational force.

The gravitational force acting on the bob is $mg$. Resolving this into components:

  • Radial component: $mg \cos\theta$ (balanced by string tension $T_{tension}$)
  • Tangential component: $F = -mg \sin\theta$ [By Newton's Second Law]

Step 2: Deriving the Effective Force Constant

For small angular displacements ($\theta \ll 1$ radian), we use the small-angle approximation $\sin\theta \approx \theta$. Since the arc length $s = l\theta$, we have $\theta = s/l$. Substituting these into the force equation:

$F = -mg \left(\frac{s}{l}\right) = -\left(\frac{mg}{l}\right)s$

Comparing this to the standard SHM force equation $F = -ks$, we identify the effective force constant $k$ for the pendulum as:

$k = \frac{mg}{l}$

θ m

Step 3: Calculating the Time Period

The general formula for the period of an oscillator is $T = 2\pi \sqrt{\frac{m}{k}}$. Substituting our derived value for $k$:

$T = 2\pi \sqrt{\frac{m}{(mg/l)}}$

Algebraic simplification yields:

$T = 2\pi \sqrt{\frac{m \cdot l}{m \cdot g}}$

$T = 2\pi \sqrt{\frac{l}{g}}$

Step 4: Logical Deduction of Mass Independence

Observe that in the expression $T = 2\pi \sqrt{\frac{l}{g}}$, the variable $m$ (mass of the bob) cancels out completely. This occurs because:

  • The restoring force ($mg\sin\theta$) is directly proportional to mass [Per Newton's Law of Universal Gravitation].
  • The inertia (resistance to acceleration) is also directly proportional to mass [Per Newton's Second Law, $F=ma$].

Since both the force driving the motion and the resistance to that motion increase linearly with mass, they cancel each other out in terms of acceleration, rendering the acceleration (and thus the period) independent of mass.

Final Solution: The mass $m$ appears in the numerator as the inertial term and in the denominator as part of the restoring force constant $k = mg/l$. Consequently, the mass terms cancel out during the derivation, proving that the time period of a simple pendulum is independent of the mass of the bob, depending only on the pendulum length and local gravitational acceleration.


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