Oscillations – NEET UG PhysicsPractice Questions & PYQs

A simple pendulum attains its maximum velocity and kinetic energy at the mean position, dropping to zero at the extreme positions where the motion momentarily pauses. Because kinetic energy is proportional to the square of velocity, defined by the equation $K = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t + \phi)$, the energy values remain strictly non-negative throughout the motion. Squaring the cosine function effectively doubles the frequency of the oscillation, implying that the kinetic energy completes two full cycles for every single time period $T$ of the pendulum. Consequently, the graphical representation of this energy appears as a waveform that never dips below the horizontal axis and displays two complete peaks within the duration of a single time period $T$.

The position of a particle executing simple harmonic motion is described by the equation $x = A \sin \left( \frac{2 \pi}{T} t \right)$, assuming it starts from equilibrium at $t = 0$. At the extreme position $x = A$, the equation becomes $A = A \sin \left( \frac{2 \pi}{T} t_1 \right)$, which implies $\sin \left( \frac{2 \pi}{T} t_1 \right) = 1$. Since the sine of $\frac{\pi}{2}$ equals $1$, the time $t_1$ is found to be $\frac{T}{4}$.

When the particle reaches $x = \frac{A}{2}$, the displacement equation becomes $\frac{A}{2} = A \sin \left( \frac{2 \pi}{T} t_2 \right)$, or $\sin \left( \frac{2 \pi}{T} t_2 \right) = \frac{1}{2}$. Recognizing that $\sin \left( \frac{\pi}{6} \right) = \frac{1}{2}$, we solve for $t_2$ to get $\frac{2 \pi}{T} t_2 = \frac{\pi}{6}$, which results in $t_2 = \frac{T}{12}$. The time interval required to travel from $x = A$ to $x = \frac{A}{2}$ is the difference between these two times, calculated as $\frac{T}{4} - \frac{T}{12}$, which simplifies to $\frac{T}{6}$.

In a simple harmonic motion system, the potential energy at any displacement $x$ is determined by the formula $U = \frac{1}{2} k x^2$, where $k$ represents the force constant. This relationship allows us to calculate the energy stored at the extreme point of motion by substituting the amplitude directly into the equation.

Given a force constant of $2 \times 10^6 \text{ N/m}$ and an amplitude of $0.01 \text{ m}$, we calculate the potential energy at this position as follows:
$U = \frac{1}{2} (2 \times 10^6 \text{ N/m}) (0.01 \text{ m})^2$
Since the square of $0.01$ is $10^{-4}$, the expression simplifies to:
$U = (1 \times 10^6) \times 10^{-4} = 100 \text{ J}$
At the amplitude, the oscillator stores 100 J of potential energy.

In a spring-mass simple harmonic oscillator, the conservation of energy dictates that the total mechanical energy remains constant and is the sum of kinetic and potential energies, expressed by the equation
$E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
Rearranging this relationship to isolate the terms on the phase plane yields
$v^2 + \frac{k}{m}x^2 = \frac{2E}{m}$
This equation typically represents an ellipse because the coefficients of the squared terms are unequal. For the graph to specifically take the form of a circle, the coefficients of both $v^2$ and $x^2$ must be identical. Since the coefficient of $v^2$ is $1$, the coefficient of $x^2$, which is $\frac{k}{m}$, must also equal $1$, leading to the condition $k = m$.

The conservation of mechanical energy dictates that the total energy of a pendulum bob remains constant, comprising both kinetic and potential components. At the equilibrium position, the height of the bob is at its lowest point, causing the potential energy to be zero; consequently, the entire total mechanical energy manifests as kinetic energy.

Since the total energy is 0.02 J, we set the kinetic energy equal to this value. Converting the mass of the bob from grams to kilograms gives 0.02 kg. Using the kinetic energy formula,
$\frac{1}{2}mv^2 = 0.02$
Substituting the known mass into this equation yields
$0.01 v^2 = 0.02$
which simplifies to
$v^2 = 2$
resulting in a speed of approximately 1.41 m/s.

In simple harmonic motion, the position of the oscillator varies as a trigonometric function, while both kinetic and potential energies are proportional to the square of these sinusoidal terms. Squaring a sine or cosine function effectively doubles its frequency, which means that the energy values complete two full cycles for every single cycle of the mass's displacement. Since the time period of the spring is 4 seconds, the period of the energy oscillations is calculated using $T' = \frac{T}{2}$. Therefore, the difference between the kinetic and potential energies varies with a period of 2 seconds.

The time period of a simple pendulum depends on its effective length, established by the relation $T = 2\pi \sqrt{\frac{L}{g}}$. Determine the time period for a single cycle by dividing the total duration of 60 s by the 30 oscillations, which equals $T = 2$ s.

To find the effective length, isolate $L$ in the formula to get $L = \frac{g T^2}{4\pi^2}$. Plugging in the values $g = 9.8$ and $\pi^2 = 9.8$ into the equation results in $L = \frac{9.8 \times 2^2}{4 \times 9.8}$. Evaluating this demonstrates that the terms simplify to a final effective length of 1 m.

When two simple harmonic motions occur simultaneously at right angles with the same angular frequency, the path of the resultant motion depends entirely on the phase difference between them. Representing these perpendicular motions as $x = a \sin \omega t$ and $y = b \sin(\omega t + \pi)$, we can see that the second expression simplifies to $y = -b \sin \omega t$. By eliminating the time parameter using the relationship $\sin \omega t = x/a$, the equation becomes $y = -(b/a)x$, which is the characteristic equation of a straight line passing through the origin.

Simple harmonic motion is fundamentally defined by a restoring force that is directly proportional to the displacement of the particle from its equilibrium position and acts in the opposite direction. Because the acceleration in this motion satisfies the condition $\frac{d^{2} x}{d t^{2}} \propto -x$, applying Newton's second law shows that the force must also be proportional to the negative of the displacement. This results in a force expression of $F = -Akx$, where the negative sign signifies that the force consistently opposes the direction of the particle's displacement to drive the oscillation.

In a forced oscillator with both a restoring force proportional to displacement and a resisting force proportional to velocity, resonance occurs when the external driving frequency matches the natural frequency of the system. Because both the amplitude and the energy of the particle reach their peaks at this point of resonance, the specific driving frequencies associated with these maxima, $\omega_{1}$ and $\omega_{2}$, are identical to the natural frequency $\omega_{0}$. This means that the condition for maximum amplitude and maximum energy is satisfied when $\omega_{1} = \omega_{0}$ and $\omega_{2} = \omega_{0}$.

The energy of a simple pendulum in harmonic motion is distributed between potential energy, given by $V = \frac{1}{2} m \omega^2 x^2$, and kinetic energy, given by $T = \frac{1}{2} m \omega^2 (a^2 - x^2)$. To determine the ratio $\frac{T}{V}$, we divide the expression for kinetic energy by the expression for potential energy. Since the factor $\frac{1}{2} m \omega^2$ appears in both the numerator and the denominator, it cancels out entirely. This simplification leaves the final ratio as $\frac{a^2 - x^2}{x^2}$.

The total mechanical energy in a simple harmonic oscillator is constant and defined by the expression $E = \frac{1}{2} m \omega^2 a^2$, where $a$ is the amplitude. At any arbitrary displacement $x$, the instantaneous kinetic energy is determined by $K = \frac{1}{2} m \omega^2 (a^2 - x^2)$. Substituting the given condition of $x = \frac{a}{2}$ into this formula allows us to identify the portion of total energy that is kinetic.

Substituting the displacement value results in $K = \frac{1}{2} m \omega^2 (a^2 - \frac{a^2}{4})$, which simplifies to $K = \frac{1}{2} m \omega^2 (\frac{3a^2}{4})$. By rearranging this expression as $K = \frac{3}{4} (\frac{1}{2} m \omega^2 a^2)$, it becomes clear that the kinetic energy is exactly three-quarters of the total energy of the system.

When a cylindrical object of uniform density floats in a liquid, its vertical oscillations are governed by the restoring buoyancy force, making the movement simple harmonic. The time period $T$ of these oscillations depends on the cylinder's dimensions and the liquid's density according to the relationship $T = 2\pi \sqrt{\frac{\rho_s l}{\rho_l g}}$, where $\rho_s$ is the cork's density and $l$ represents its length. Because the cork and its dimensions are identical in both scenarios, the time period is inversely proportional to the square root of the liquid's density, expressed as $T \propto \frac{1}{\sqrt{\rho_l}}$. Comparing the two cases where $T_1 = T$ and $T_2 = 2T$, we establish the ratio $\frac{T}{2T} = \sqrt{\frac{\rho_2}{\rho_1}}$, which simplifies to $\frac{1}{2} = \sqrt{\frac{\rho_2}{\rho_1}}$. Squaring both sides of this relationship reveals that $\frac{\rho_2}{\rho_1} = \frac{1}{4}$.

The total mechanical energy $E$ of a body undergoing simple harmonic motion is defined by the expression $E = \frac{1}{2} M \omega^{2} A^{2}$, while the potential energy at any displacement $x$ from the mean position is given by $\mathrm{P.E} = \frac{1}{2} M \omega^{2} x^{2}$. To find the specific displacement where the potential energy is exactly one-fourth of the total energy, we set the potential energy equal to $\frac{1}{4} E$:

$\frac{1}{2} M \omega^{2} x^{2} = \frac{1}{4} \left( \frac{1}{2} M \omega^{2} A^{2} \right)$

Canceling the common term $\frac{1}{2} M \omega^{2}$ from both sides leads to the simplified relation $x^{2} = \frac{A^{2}}{4}$, which confirms that the required displacement is $x = \frac{A}{2}$.

In simple harmonic motion, the maximum speed of an oscillating object is governed by the product of its amplitude and angular frequency, expressed as $V = A \omega$. Equating the maximum speeds of the two systems, $A_P \omega_P = A_Q \omega_Q$, reveals that the amplitude ratio is inversely proportional to the frequency ratio, $\frac{A_Q}{A_P} = \frac{\omega_P}{\omega_Q}$. Since the angular frequency for a spring-mass system is defined by $\omega = \sqrt{k/m}$ and both masses are identical, substituting these values allows us to write the ratio as $\frac{\sqrt{k_1/m}}{\sqrt{k_2/m}}$. Canceling the common mass terms results in the final ratio:

$\frac{A_Q}{A_P} = \sqrt{\frac{k_1}{k_2}}$

In a spring-mass system, the time period of oscillation is governed by the equation
$T = 2\pi\sqrt{\frac{m}{k}}$
As sand leaks from the box, the mass $m$ decreases, which causes the time period $T$ to decrease over time. Since the angular frequency is related to the time period by $\omega = \frac{2\pi}{T}$, the decreasing time period results in a corresponding increase in the angular frequency $\omega$. Finally, because the equilibrium displacement is defined by $x_0 = \frac{mg}{k}$, the continuous loss of mass causes the equilibrium position to shift upward, which results in the oscillation amplitude $A$ decreasing over time.

The oscillation period of a simple pendulum depends exclusively on the length of the string and the gravitational acceleration, meaning the mass of the bob has no effect on the timing. Since the initial time period is defined by $T = 2\pi\sqrt{\frac{\ell}{g}}$, substituting the new length $\ell' = \frac{\ell}{2}$ results in an adjusted period $T' = 2\pi\sqrt{\frac{\ell/2}{g}}$. This expression simplifies to $T' = \frac{1}{\sqrt{2}} \left( 2\pi\sqrt{\frac{\ell}{g}} \right)$, which shows that the new period is equal to $\frac{1}{\sqrt{2}}T$. Given the problem defines the new period as $\frac{x}{2}T$, we equate $\frac{x}{2} = \frac{1}{\sqrt{2}}$, which leads to $x = \frac{2}{\sqrt{2}}$, confirming that $x = \sqrt{2}$.

The position of the particle is defined by the vector equation $\vec{r}=(\hat{i}+2\hat{j})A\cos\omega t$, which decomposes into the scalar components $x=A\cos\omega t$ and $y=2A\cos\omega t$. Eliminating the time variable reveals the relationship $y=2x$, indicating that the particle traces a straight-line path rather than a parabolic or elliptical one. Differentiating the position vector twice with respect to time yields an acceleration of $\vec{a}=-\omega^2(\hat{i}+2\hat{j})A\cos\omega t$, which simplifies to the characteristic form $\vec{a}=-\omega^2\vec{r}$ required for simple harmonic motion. Because the position vector repeats at regular intervals as defined by the cosine function, the movement is also classified as periodic.

In simple harmonic motion, the total mechanical energy is conserved and oscillates between kinetic energy and potential energy depending on the particle's displacement from the equilibrium position. The potential energy is given by the relation
$U = \frac{1}{2} k x^2$
while the kinetic energy is expressed as
$K = \frac{1}{2} k (A^2 - x^2)$
To find the displacement where these two energies are equal, we set the expressions against each other:
$\frac{1}{2} k x^2 = \frac{1}{2} k (A^2 - x^2)$
Canceling the common terms simplifies the equation to $x^2 = A^2 - x^2$, which rearranges to $2x^2 = A^2$. Solving for displacement $x$ yields $x = \frac{A}{\sqrt{2}}$.

Simple Harmonic Motion is described by the general displacement equation $x = A \sin(\omega t + \phi)$, where $A$ represents the amplitude and $\omega$ represents the angular frequency. By aligning this with the expression $x = 5 \sin(\pi t + \frac{\pi}{3})$ m, the amplitude is identified as 5 m. Since the angular frequency equals $\pi$, the time period $T$ is found using the fundamental relationship $\omega = \frac{2\pi}{T}$. Substituting the known value of $\omega$ into this formula leads to:

$T = \frac{2\pi}{\pi} = 2$

The time period is 2 s.