Class 11 Physics (Physics Part-II) Chapter 13 Oscillations

Key Concept: SHM Characteristics

d) The force is proportional to displacement and directed toward the mean position.

[Solution Description] In Simple Harmonic Motion (SHM), by definition, the force acting on the object is directly proportional to its displacement from the equilibrium position and is directed towards that position. This restoring nature of the force ensures oscillatory behavior.

Key Concept: Basic Formula

c) $T = 2\pi \sqrt{\frac{L}{g}}$
[Solution Description] The time period $T$ of a simple pendulum, which depends on its length and the acceleration due to gravity, is given by the formula:
$T = 2\pi \sqrt{\frac{L}{g}}$
This formula shows that the time period $T$ is proportional to the square root of the ratio of the length $L$ of the pendulum to the gravitational acceleration $g$.

Key Concept: Velocity Zero Points

b) At maximum displacement

[Solution Description] In SHM, the velocity of the particle becomes zero at the maximum displacement or amplitude since the particle changes direction at these points.

Key Concept: Period Calculation

a) $\frac{1}{2}$ s
[Solution Description] The period $T$ can be calculated using the formula $T = \frac{2\pi}{\omega}$. Substituting $\omega = 4\pi$, we get $T = \frac{2\pi}{4\pi} = \frac{1}{2}$ s.

Key Concept: Real-world Application, Displacement Variables

d) Amplitude of oscillation gradually decreases.
[Solution Description]
In a real-world application of SHM involving damping, the motion becomes underdamped or overdamped based on the magnitude of the damping coefficient $b$. The equation of motion for a damped oscillator is modified from $F = -kx$ to $F = -kx – bv$. If voltage across a capacitor represents displacement, its amplitude decreases exponentially over time due to energy loss in the form of heat, unlike in ideal SHM where amplitude remains constant.

Key Concept: Energy Conservation

b) Kinetic Energy

[Solution Description]

At the equilibrium position in simple harmonic motion (SHM), the displacement from the mean position is zero. Here, the entire energy of the system is converted into kinetic energy, making it maximal. Therefore, at equilibrium position, the kinetic energy is at its maximum while the potential energy is zero.

Key Concept: Force Formula

b) $F(t) = -m\omega^2 x(t)$
[Solution Description] In simple harmonic motion, the force acting on a particle is given by Newton’s second law applied to SHM. The force $F(t)$ can be expressed as:
$F(t) = -m\omega^2 x(t)$
where $m$ is the mass of the particle, $\omega$ is the angular frequency, and $x(t)$ is the displacement.
This equation reflects that the force is proportional to the displacement and directed towards the equilibrium position.

Key Concept: Basic Concept

b) mg sinθ
[Solution Description] The restoring force in a simple pendulum is provided by the tangential component of gravity, which can be expressed as $F_{\text{restoring}} = mg \sinθ$. This force acts to bring the pendulum back towards its equilibrium position.

Key Concept: Potential Energy

b) At the extreme positions
[Solution Description] In simple harmonic motion, the potential energy is represented by the formula:
$U = \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \cos^2(\omega t + \phi)$
The potential energy is maximum when $\cos^2(\omega t + \phi)$ is maximum, which happens at the extreme positions $x = \pm A$.

Key Concept: Energy Calculation

b) 1 J
[Solution Description]
The potential energy in simple harmonic motion is calculated using the formula $U(x) = \frac{1}{2}kx^2$.
Let’s substitute the given values:
$U(x) = \frac{1}{2} \times 200 \, \text{N/m} \times (0.1 \, \text{m})^2$
Simplifying this:
$U(x) = 100 \, \text{N/m} \times 0.01 \, \text{m}^2 = 1 \, \text{J}$
Thus, the potential energy is $1 \, \text{J}$.

Key Concept: Equilibrium Position

b) At the lowest point

[Solution Description] The equilibrium position of a simple pendulum is where the pendulum comes to rest when not in motion. It is typically at the lowest point of its swing.

Key Concept: Conceptual Understanding

b) The smallest time interval after which motion repeats

[Solution Description] The period of a periodic motion is the smallest interval of time after which the motion repeats itself. It represents the duration for one complete cycle.

Key Concept: Force and Displacement

d) Force is proportional to displacement
[Solution Description]
In simple harmonic motion, the restoring force is directly proportional to the negative of the displacement from the equilibrium position. This means if displacement increases, the force also increases but in the opposite direction. Mathematically, it can be represented as $F = -kx$, where $k$ is the spring constant and $x$ is the displacement. Thus, the correct statement is that force is proportional to displacement.

Key Concept: Basic Acceleration

d) $a(t) = -\omega^2 A \cos(\omega t + \phi)$
[Solution Description]
The acceleration $a(t)$ of a particle in SHM is given by:
$a(t) = -\omega^2 A \cos(\omega t + \phi)$
where $\omega$ is the angular frequency, $A$ is the amplitude, and $\phi$ is the phase angle.

Key Concept: Simple Definitions

b) Amplitude

[Solution Description] In oscillatory motion, amplitude is defined as the maximum displacement of the object from its mean or equilibrium position.

Key Concept: Basic Definition

c) Periodic

[Solution Description] A simple pendulum exhibits periodic motion, meaning it moves back and forth in a regular repeating cycle due to gravitational force.

Key Concept: Basic Concept

d) A small mass and a string

[Solution Description] The simple pendulum consists of a small mass known as the bob, which is attached to a string that allows it to swing freely in an arc.

Key Concept: Mathematical Function

d) Both a) and b)
[Solution Description]
For a function to be periodic, it must repeat itself at regular intervals over time. Both sine and cosine functions are periodic with period $2\pi$. The given options include:
\begin{align*}
\text{a) } f(t) &= A \sin(\omega t)
\text{b) } f(t) &= A \cos(\omega t)
\text{c) } f(t) &= A \tan(\omega t)
\text{d) } f(t) &= A \cot(\omega t)
\end{align*}
Among these, $f(t) = A \sin(\omega t)$ and $f(t) = A \cos(\omega t)$ are periodic. Thus, both A and B qualify as periodic functions.

Key Concept: Basic Formula

b) $a(t) = -\omega^2 A \cos(\omega t + \phi)$
[Solution Description]
The formula for acceleration $a(t)$ in SHM is based on the displacement function $x(t) = A \cos(\omega t + \phi)$. Acceleration is derived as the second derivative of the displacement with respect to time.
Given:
$a(t) = -\omega^2 A \cos(\omega t + \phi)$
This formula shows that the acceleration is directly proportional to the negative of the displacement with an additional factor of $\omega^2$.

Key Concept: Basic Formula

d) $K = \frac{1}{2} k A^2 \sin^2(\omega t + \phi)$
[Solution Description] The kinetic energy ($K$) in simple harmonic motion can be calculated using the formula $K = \frac{1}{2} m v^2$. Expressing this in terms of amplitude and angular frequency, we have:
$K = \frac{1}{2} k A^2 \sin^2(\omega t + \phi)$
where $m$ is mass, $v$ is velocity, $k$ is the spring constant, $A$ is amplitude, $\omega$ is angular frequency, and $\phi$ is phase angle.

Key Concept: Definition of SHM

c) $F(t) = -k x(t)$
[Solution Description]
The definition of simple harmonic motion in terms of force is given by the equation $F(t) = -k x(t)$, which shows that the force acting on the particle is directly proportional and opposite to the displacement from its mean position.

Key Concept: Period and Frequency

b) 0.5 Hz
[Solution Description]
The period $T$ of the pendulum is given as 2 seconds. The frequency $f$ is the reciprocal of the period, calculated using the formula $f = \frac{1}{T}$. Hence, the frequency is $f = \frac{1}{2} = 0.5 \text{ Hz}$.

Key Concept: Energy Variation

b) Kinetic energy is greater than potential energy.
[Solution Description]
The total energy in SHM is constant and given by:
$E = \frac{1}{2} k A^2$
Potential energy at a displacement $x = \frac{A}{2}$ is:
$U = \frac{1}{2} k \left(\frac{A}{2}\right)^2 = \frac{1}{8} k A^2$
Therefore, the kinetic energy at this displacement is:
$K = E – U = \frac{1}{2} k A^2 – \frac{1}{8} k A^2 = \frac{3}{8} k A^2$
Thus, the kinetic energy is greater than the potential energy at this point, which is option b).

Key Concept: Basic Formula Application

a) $F(t) = -kx(t)$
[Solution Description] The force acting on a particle in simple harmonic motion can be given by either $F(t) = -kx(t)$ or $F(t) = -m\omega^2 x(t)$. Both formulas represent the same concept that the force is proportional to the displacement from the mean position.

Key Concept: Velocity in SHM

b) $v(t) = -\omega A \sin(\omega t + \phi)$
[Solution Description] To find the velocity, we differentiate the displacement equation $x(t) = A \cos(\omega t + \phi)$ with respect to time $t$.
The derivative of $\cos(\omega t + \phi)$ is $-\sin(\omega t + \phi)$ and applying chain rule gives us an additional factor of $\omega$. Thus,
$v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi)$
Hence, the velocity is $v(t) = -\omega A \sin(\omega t + \phi)$.

Key Concept: Basic Formula Application

c) $18 m/s^2$
[Solution Description] The maximum acceleration $a_{\text{max}}$ in SHM is given by $a_{\text{max}} = \omega^2 A$. Substituting $\omega = 3 \, \text{rad/s}$ and $A = 2 \, \text{m}$ into the formula:
$a_{\text{max}} = (3)^2 \times 2 = 9 \times 2 = 18 \, \text{m/s}^2$
So, the maximum acceleration is $18 m/s^2$.

Key Concept: Oscillation vs Vibration

b) Frequency

[Solution Description]

Oscillation and vibration are different based on their frequency. Generally, low-frequency motions are termed oscillations, while high-frequency motions are termed vibrations.

Key Concept: Angular Frequency Calculation

b) $\frac{2\pi}{5}$ rad/s
[Solution Description]
Angular frequency $\omega$ is related to the period $T$ by the formula $\omega = \frac{{2\pi}}{{T}}$. Given that the period $T = 5$ seconds:
$\omega = \frac{{2\pi}}{{5}}$
Therefore, the angular frequency is $\frac{{2\pi}}{{5}}$ rad/s.

Key Concept: Complex Phase Analysis, Real-world Application

b) $-\frac{\pi}{3}$
[Solution Description] Given that the displacement $x(0) = \frac{A}{2}$, using the equation $x(t) = A \cos(\omega t + \phi)$, we have $A \cos(\phi) = \frac{A}{2}$. Dividing both sides by $A$, we get $\cos(\phi) = \frac{1}{2}$, thus $\phi = \pm \frac{\pi}{3}$. Since the velocity is towards the mean position at $t=0$, we use the negative sign for $\sin(\phi)$. Therefore, $\phi = -\frac{\pi}{3}$.

Key Concept: Complex Projection Analysis, Real-world Application

c) $\pi/2$
[Solution Description] The problem involves analyzing the relationship between two SHMs that are perpendicular projections of the same circular motion. For such projections, we consider the general form: $x(t) = R \cos(\omega t + \phi_1)$ and $y(t) = R \sin(\omega t + \phi_2)$. Since both are projections from the same circular motion, their phase difference should account for the perpendicular nature, which implies a phase difference of $\pi/2$ radians.

Key Concept: Periodicity

c) $T = \frac{2\pi}{\omega}$
[Solution Description] The period $T$ of simple harmonic motion is related to the angular frequency $\omega$ by the equation $\omega = \frac{2\pi}{T}$. Solving for $T$, we get $T = \frac{2\pi}{\omega}$.

Key Concept: Equilibrium Concept

b) It performs oscillations

[Solution Description] When an object is displaced from its equilibrium position in oscillatory motion, it experiences a restoring force that causes it to return towards equilibrium, causing it to perform oscillations around that point until external forces or damping eventually bring it to rest.

Key Concept: Phase Difference, Complex Problem Solving

c) $\pi$
[Solution Description]
In a two-block system connected by a spring, both blocks are subject to equal and opposite forces. Therefore, their motions are out of phase by $\pi$ radians or $180^\circ$. If the displacement of one block is described as $x_1(t) = A \cos(\omega t + \phi)$, then for the other block moving oppositely, the displacement can be written as $x_2(t) = A \cos(\omega t + \phi + \pi)$, resulting in a phase difference of $\pi$.

Key Concept: Period and Frequency Relationship

b) 0.25 Hz
[Solution Description] Frequency $\nu$ is the reciprocal of the period $T$. So, if the period $T = 4$ seconds, then the frequency $\nu = \frac{1}{T} = \frac{1}{4}$ Hz or 0.25 Hz.