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For the following questions, statement as well as the reason(s) are given. Each questions has four options. Select the correct option. (a) Statement \(-1\) is true, statement \(-2\) is true; statement \(-2\) is the correct explanation of statement \(-1\) (b) Statement \(-1\) is true, statement \(-2\) is true but statement \(-2\) is not the correct explanation of statement \(-1\) (c) Statement \(-1\) is true, statement \(-2\) is false (d) Statement \(-1\) is false, statement \(-2\) is true (A) a (B) \(b\) (C) \(\mathrm{c}\) (D) \(\mathrm{d}\) Statement \(-1:\) Periodic time of a simple pendulum is independent of the mass of the bob. Statement \(-2:\) The restoring force does not depend on the mass of the bob. (A) a (B) \(b\) (C) \(\mathrm{c}\) (D) \(\mathrm{d}\)

Short Answer

Expert verified
(C) Statement \(-1\) is true, statement \(-2\) is false

Step by step solution

01

Analyze Statement 1

The periodic time of a simple pendulum is given by the formula \(T = 2\pi \sqrt{\frac {L}{g}}\), where \(T\) is the period, \(L\) is the length of the pendulum, and \(g\) is the acceleration due to gravity. As we can see from the formula, the period does not depend on the mass of the bob. Therefore, Statement 1 is true.
02

Analyze Statement 2

The equation for the restoring force acting on a pendulum bob in a simple pendulum system is \(F = -mgsin(\theta)\), where \(F\) is the restoring force, \(m\) is the mass of the bob, \(g\) is the acceleration due to gravity, and \(\theta\) is the angular displacement of the pendulum from its equilibrium position. As we can see from the equation, the restoring force does depend on the mass of the bob. Therefore, Statement 2 is false.
03

Check Explanation

As Statement 1 is true and Statement 2 is false, there is no need to check if Statement 2 explains Statement 1 since Statement 2 is not the correct explanation for Statement 1. Based on our analysis, the correct option is: (C) Statement \(-1\) is true, statement \(-2\) is false

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Periodic Time
The periodic time of a simple pendulum refers to the time it takes for the pendulum to complete one full swing, going from one side back to the same point again. This interval is very important because it allows us to know how long the pendulum takes to oscillate once. In physics, the formula that determines the periodic time of a simple pendulum is given by:\[ T = 2\pi \sqrt{\frac {L}{g}} \]where:
  • \( T \) is the periodic time.
  • \( L \) is the length of the pendulum.
  • \( g \) is the acceleration due to gravity.
The equation reveals that the periodic time does not depend on the mass of the bob. This fascinating aspect makes the simple pendulum a great tool for teaching physics. It shows that characteristics such as the length of the pendulum and gravitational pull affect the period, while the mass does not.
This makes it easier to predict the pendulum's behavior without needing to change the bob's mass.
Restoring Force
In a simple pendulum, the restoring force is the force that tries to bring the pendulum back to its equilibrium position. This force is what makes the pendulum swing back when it is displaced from its resting state. According to physics, the restoring force is calculated using the formula:\[ F = -mg\sin(\theta) \]Where:
  • \( F \) is the restoring force.
  • \( m \) is the mass of the bob.
  • \( g \) is acceleration due to gravity.
  • \( \theta \) is the angle of displacement.
From the equation, it's clear that the restoring force does rely on the mass of the bob, as well as the angle of displacement. The negative sign indicates that the force is directed opposite to the displacement, helping to stabilize the pendulum's motion. This interaction creates the oscillation characteristic of pendulum movement.
Mass Independence
A particularly interesting aspect of a simple pendulum is its mass independence, specifically its periodic time being independent of the bob's mass. Many other systems, such as springs or waves, will depend on the mass involved. However, in simple pendulums, the periodic time formula:\[ T = 2\pi \sqrt{\frac {L}{g}} \]shows no reference to mass, which implies that changing the mass of the pendulum bob does not affect the time it takes to complete a swing.
This aspect is counterintuitive because one might think a heavier object would take longer. Yet, due to gravitational influence and inertia, changing the mass doesn't alter the periodic time. This phenomenology is captivating for students and educators alike, as it underscores the unique behaviors of pendulums and introduces learners to fundamental physics concepts without the complexity of mass variables.

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Most popular questions from this chapter

A wave travelling along a string is described by \(\mathrm{y}=0.005 \sin (40 \mathrm{x}-2 \mathrm{t})\) in SI units. The wavelength and frequency of the wave are \(\ldots \ldots \ldots\) (A) \((\pi / 5) \mathrm{m} ; 0.12 \mathrm{~Hz}\) (B) \((\pi / 10) \mathrm{m} ; 0.24 \mathrm{~Hz}\) (C) \((\pi / 40) \mathrm{m} ; 0.48 \mathrm{~Hz}\) (D) \((\pi / 20) \mathrm{m} ; 0.32 \mathrm{~Hz}\)

An listener is moving towards a stationary source of sound with a speed (1/4) times the speed of sound. What will be the percentage increase in the frequency of sound heard by the listener? (A) \(20 \%\) (B) \(25 \%\) (C) \(2.5 \%\) (D) \(5 \%\)

A tuning fork arrangement produces 4 beats/second with one fork of frequency \(288 \mathrm{~Hz}\). A little wax is applied on the unknown fork and it then produces 2 beats/s. The frequency of the unknown fork is \(\ldots \ldots \ldots . \mathrm{Hz}\). (A) 286 (B) 292 (C) 294 (D) 288

A block having mass \(\mathrm{M}\) is placed on a horizontal frictionless surface. This mass is attached to one end of a spring having force constant \(\mathrm{k}\). The other end of the spring is attached to a rigid wall. This system consisting of spring and mass \(\mathrm{M}\) is executing SHM with amplitude \(\mathrm{A}\) and frequency \(\mathrm{f}\). When the block is passing through the mid-point of its path of motion, a body of mass \(\mathrm{m}\) is placed on top of it, as a result of which its amplitude and frequency changes to \(\mathrm{A}^{\prime}\) and \(\mathrm{f}\). The ratio of amplitudes \(\left(\mathrm{A}^{1} / \mathrm{A}\right)=\ldots \ldots \ldots\) (A) \(\sqrt{\\{}(\mathrm{M}+\mathrm{m}) / \mathrm{m}\\}\) (B) \(\sqrt{\\{m} /(\mathrm{M}+\mathrm{m})\\}\) (C) \(\sqrt{\\{} \mathrm{M} /(\mathrm{M}+\mathrm{m})\\}\) (D) \(\sqrt{\\{}(\mathrm{M}+\mathrm{m}) / \mathrm{M}\\}\)

A tuning fork of frequency \(480 \mathrm{~Hz}\) produces 10 beats/s when sounded with a vibrating sonometer string. What must have been the frequency of the string if a slight increase in tension produces fewer beats per second than before? (A) \(480 \mathrm{~Hz}\) (B) \(490 \mathrm{~Hz}\) (C) \(460 \mathrm{~Hz}\) (D) \(470 \mathrm{~Hz}\)

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