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Question

Which of the following statements is false with respect to a simple pendulum?

The correct answer is

The time period of simple pendulum is directly proportional to mass of the body suspended at the free end of string.

A simple pendulum is a basic mechanical system consisting of a point mass (often called a bob) suspended from a fixed point by a light, inextensible string. When displaced from its equilibrium position, the simple pendulum swings back and forth due to gravity, performing oscillatory motion. The time period of a simple pendulum is the time taken for one complete oscillation (i.e., for the pendulum to swing out and return to its starting position).

Pendulum Time Period Formula

The time period (\(T\)) of a simple pendulum, for small angular displacements, is given by the formula:

\[T = 2\pi\sqrt{\frac{L}{g}}\]

Where:

  • \(T\) is the time period of the simple pendulum.
  • \(L\) is the effective length of the string (distance from the point of suspension to the center of mass of the bob).
  • \(g\) is the acceleration due to gravity at the location of the pendulum.

From this formula, we can analyze the dependence of the time period on various factors. Noticeably, the mass of the bob (\(m\)) is not present in this formula.

Time Period Statement Analysis

Let's examine each statement given in the context of the simple pendulum time period:

  • Statement 1: "The time period of simple pendulum is directly proportional to mass of the body suspended at the free end of string."

    This statement is false. As seen from the formula \(T = 2\pi\sqrt{\frac{L}{g}}\), the time period (\(T\)) does not depend on the mass of the body (bob) suspended. In ideal conditions, a heavier bob and a lighter bob will have the same time period if their lengths and the local acceleration due to gravity are identical. This is a crucial characteristic of a simple pendulum.

  • Statement 2: "The time period of simple pendulum is directly proportional to square root of length of the string."

    This statement is true. From the formula, we can see \(T \propto \sqrt{L}\). This means if you increase the length of the string, the time period will increase, making the pendulum swing slower. For example, if you quadruple the length, the time period will double.

  • Statement 3: "The time period of simple pendulum is inversely proportional to square root of acceleration due to gravity."

    This statement is true. From the formula, we can see \(T \propto \frac{1}{\sqrt{g}}\). This implies that if the acceleration due to gravity increases (e.g., on a planet with stronger gravity), the time period will decrease, causing the pendulum to swing faster.

  • Statement 4: "The time period of simple pendulum does not depend on mass of the body suspended at the end of string."

    This statement is true. This directly confirms what we observed from the formula: the mass of the body is not a factor influencing the time period of an ideal simple pendulum.

Conclusion on Simple Pendulum Characteristics

Based on the analysis, the statement that is false with respect to a simple pendulum is that its time period is directly proportional to the mass of the body suspended. The time period is primarily governed by the length of the string and the acceleration due to gravity.

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Important Questions from Simple Mass System

  1. A flexible rotor-shaft system comprises of a 10 kg rotor disc placed in the middle of a massless shaft of diameter 30 mm and length 500 mm between bearings (shaft is being taken mass-less as the equivalent mass of the shaft is included in the rotor mass) mounted at the ends. The bearings are assumed to simulate simply supported boundary conditions. The shaft is made of steel for which the value of E is 2.1 x 1011 Pa. What is the critical speed of rotation of the shaft?

  2. Natural frequency (ωn) of a passenger car whose weight is w Newton and whose suspension has a combined stiffness of k N/mm is given by:

  3. If mass M oscillates on a spring having mass m and stiffness k, then the natural frequency of the system is

  4. A simple spring mass vibrating system has a natural frequency of fn. If the spring stiffness is halved and mass is double, then the natural frequency will become

  5. The equation of motion for a spring-mass system excited by a harmonic force is

    \(M\ddot x + kx = F\cos \left( {\omega t} \right),\)

    Where M is the mass, K is the spring stiffness, F is the force amplitude and ω is the angular frequency of excitation. Resonance occurs when ω is equal to
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