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A simple pendulum having length \(\ell\) is given a small angular displacement at time \(t=0\) and released. After time \(t\), the linear displacement of the bob of the pendulum is given by \(\ldots \ldots \ldots \ldots\) (A) \(x=a \sin 2 p \sqrt{(\ell / g) t}\) (B) \(\mathrm{x}=\mathrm{a} \cos 2 \mathrm{p} \sqrt{(\mathrm{g} / \ell) t}\) (C) \(\mathrm{x}=\mathrm{a} \sin \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\) (D) \(\mathrm{x}=\mathrm{a} \cos \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\)

Short Answer

Expert verified
The short version of the answer is: \(x(t) = a \cos(\sqrt{\frac{g}{\ell}t})\) The correct option is (D).

Step by step solution

01

Establish the relationship between angular displacement and linear displacement.

The linear displacement (x) of the bob of the pendulum is related to the angular displacement (θ) as follows: x = ℓ * θ
02

Write the equation for angular displacement as a function of time.

For a simple pendulum with small angular displacements, the angular displacement θ(t) as a function of time can be expressed using the equation: θ(t) = θ₀ * cos(ωt) where θ₀ is the initial angular displacement, ω is the angular frequency, and t is the time.
03

Calculate the angular frequency (ω).

The angular frequency (ω) of a simple pendulum is given by the formula: ω = √(g/ℓ) where g is the acceleration due to gravity and ℓ is the length of the pendulum.
04

Substitute the angular frequency (ω) and angular displacement (θ) into the equation for linear displacement (x).

We now have everything needed to calculate the linear displacement x(t) as a function of time. Substitute θ(t) and ω into the equation for x: x(t) = ℓ * θ₀ * cos(√(g/ℓ)t)
05

Simplify the equation for linear displacement.

To simplify the equation for linear displacement, let a = ℓ * θ₀: x(t) = a cos(√(g/ℓ)t)
06

Compare with given options.

The final equation for linear displacement as a function of time is: x(t) = a cos(√(g/ℓ)t) Comparing this equation with the given options, we find that the correct answer is: (D) x = a cos √((g/ℓ)t)

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Key Concepts

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

Linear Displacement
Linear displacement in a simple pendulum represents the distance the pendulum bob moves from its equilibrium position along the arc. It's closely related to angular displacement, which is a description of the pendulum's position in terms of angles.
To find linear displacement, we use the relationship between linear displacement (\( x \) ) and angular displacement (\( \theta \)): \( x = \ell \cdot \theta \). Here, \( \ell \) is the pendulum length, and \( \theta \) is the angle in radians from the vertical equilibrium position.
This means the linear displacement is the arc length covered by the pendulum bob. By understanding how angular motions translate into linear motions, we can effectively describe the dynamics of pendulum systems.
Angular Displacement
Angular displacement in a pendulum is the angle through which the pendulum moves from its rest or equilibrium position. Think of it as the angle made by the pendulum arm with the vertical line.
In our context, it is typically denoted by \( \theta \), and its motion can be modeled by simple harmonic motion equations. For small angles, angular displacement varies over time following\( \theta(t) = \theta_0 \cos(\omega t) \), where \( \theta_0 \) is the initial angle, and \( \omega \) is the angular frequency.
Angular displacement reveals how oscillations occur over time, offering a way to predict the pendulum's position at any given moment based on its oscillatory nature.
Angular Frequency
Angular frequency is fundamental in describing the rate of oscillation of a pendulum. Denoted by \( \omega \), it defines how quickly the pendulum swings back and forth.
For a simple pendulum, the formula is:\[ \omega = \sqrt{\frac{g}{\ell}} \]
Here, \( g \) is gravity's acceleration, and \( \ell \) is the pendulum’s length. Angular frequency plays a crucial role in harmonic motion equations, influencing how fast and how many times in a given period the pendulum bob swings.
Understanding angular frequency helps us predict pendulum cycles, enabling an accurate calculation of oscillation patterns, periods, and frequencies for various pendulum lengths and gravitational forces.
Small Angle Approximation
The small angle approximation is a helpful simplification in physics, particularly for pendulums. When the pendulum's angular displacement is small, the sine of the angle is almost equal to the angle itself when measured in radians:
\( \sin(\theta) \approx \theta \)
This approximation allows us to use simple harmonic motion equations to describe the pendulum's movement, as it simplifies calculations significantly. It is valid when the angle \( \theta \) is less than about 15 degrees.
With this approximation, complex trigonometric equations become manageable, offering an efficient way to analyze pendulum motion. Ultimately, this simplification is essential for understanding the foundational behavior of pendulums in classical mechanics.

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Most popular questions from this chapter

Two masses \(m_{1}\) and \(m_{2}\) are attached to the two ends of a massless spring having force constant \(\mathrm{k}\). When the system is in equilibrium, if the mass \(\mathrm{m}_{1}\) is detached, then the angular frequency of mass \(m_{2}\) will be \(\ldots \ldots \ldots .\) (A) \(\sqrt{\left(\mathrm{k} / \mathrm{m}_{1}\right)}\) (B) \(\sqrt{\left(\mathrm{k} / \mathrm{m}^{2}\right)}\) (C) \(\sqrt{\left(k / m_{2}\right)+m_{1}}\) (D) \(\sqrt{\left\\{k /\left(m_{1}+m_{2}\right)\right\\}}\)

A spring is attached to the center of a frictionless horizontal turn table and at the other end a body of mass \(2 \mathrm{~kg}\) is attached. The length of the spring is \(35 \mathrm{~cm}\). Now when the turn table is rotated with an angular speed of \(10 \mathrm{rad} \mathrm{s}^{-1}\), the length of the spring becomes \(40 \mathrm{~cm}\) then the force constant of the spring is \(\ldots \ldots \mathrm{N} / \mathrm{m}\). (A) \(1.2 \times 10^{3}\) (B) \(1.6 \times 10^{3}\) (C) \(2.2 \times 10^{3}\) (D) \(2.6 \times 10^{3}\)

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 frequencies \((\mathrm{f} / \mathrm{f})=\ldots \ldots \ldots\) (A) \(\sqrt{\\{} \mathrm{M} /(\mathrm{m}+\mathrm{M})\\}\) (B) \(\sqrt{\\{\mathrm{m} /(\mathrm{m}+\mathrm{M})\\}}\) (C) \(\sqrt{\\{\mathrm{MA} / \mathrm{mA}}\\}\) (D) \(\sqrt{[}\\{(\mathrm{M}+\mathrm{m}) \mathrm{A}\\} / \mathrm{mA}]\)

The equation for displacement of a particle at time \(t\) is given by the equation \(\mathrm{y}=3 \cos 2 \mathrm{t}+4 \sin 2 \mathrm{t}\). The motion of the particle is \(\ldots \ldots\) (A) Damped motion (B) Periodic motion (C) Rotational motion (D) S.H.M.

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) \(\mathrm{d}\)

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