Oscillations – NEET UG PhysicsPractice Questions & PYQs

When a simple pendulum is submerged in a non-viscous liquid, the buoyant force exerts an upward force on the bob, which opposes gravity and effectively reduces the downward pull. This modification changes the acceleration due to gravity experienced by the pendulum to an effective value of $g_{\text{eff}} = g \left(1 - \frac{\rho}{\sigma}\right)$, where $\rho$ is the density of the liquid and $\sigma$ is the density of the material of the bob. Because the time period of a simple pendulum is inversely proportional to the square root of the effective gravity, the ratio of the new time period in the liquid to the original time period in air can be expressed as follows:

$T_{\text{liq}} = \frac{T_{\text{air}}}{\sqrt{1 - \frac{\rho}{\sigma}}}$

Given that the density of the liquid is one-fourth the density of the bob, the ratio $\frac{\rho}{\sigma}$ is $\frac{1}{4}$. Substituting the original time period of $\sqrt{3} \text{ s}$ into this relationship yields the new period:

$T_{\text{liq}} = \frac{\sqrt{3}}{\sqrt{1 - \frac{1}{4}}} = \frac{\sqrt{3}}{\sqrt{\frac{3}{4}}} = \frac{\sqrt{3}}{\frac{\sqrt{3}}{2}} = 2 \text{ s}$

The position of a particle in simple harmonic motion is described by the equation $x = A \sin(\omega t)$, where the graph indicates an amplitude $A$ of 1 and a period of 8 s, yielding an angular frequency $\omega = \frac{2\pi}{8}$ or $\frac{\pi}{4}$ rad/s. Because the acceleration relates to position as $a = -\omega^2 x$, we derive the expression $a = -(\frac{\pi}{4})^2 \sin(\frac{\pi}{4} t)$, which simplifies to $a = -\frac{\pi^2}{16} \sin(\frac{\pi}{4} t)$. Evaluating this equation at $t = 2$ s results in $a = -\frac{\pi^2}{16} \sin(\frac{\pi}{2})$, leaving a final acceleration of $-\frac{\pi^2}{16}$ m s$^{-2}$.

The time period of a simple pendulum is proportional to the square root of its length, defined by $T = 2\pi\sqrt{\frac{L}{g}}$. Two pendulums return to the same phase when the total time elapsed is equal, satisfying the condition $n_1 T_1 = n_2 T_2$, where $n_1$ and $n_2$ represent the number of full oscillations for the longer and shorter pendulums respectively. By calculating the ratio of the time periods for the given lengths of $121 \,\text{cm}$ and $100 \,\text{cm}$, we find $\frac{T_1}{T_2} = \sqrt{\frac{121}{100}} = \frac{11}{10}$. Since $n_1 T_1 = n_2 T_2$, it follows that the ratio of oscillations must be inversely proportional to the periods, such that $\frac{n_2}{n_1} = \frac{T_1}{T_2} = \frac{11}{10}$. To satisfy this condition using the smallest integer values, the shorter pendulum must complete 11 vibrations.

Periodic motion occurs when a function repeats its value at fixed intervals, a characteristic shared by all harmonic expressions involving sine and cosine. Because trigonometric functions oscillate indefinitely, the expressions $\sin(\omega t + \frac{\pi}{4})$, $\sin(\omega t)$, and $\sin(\omega t) + \cos(\omega t)$ all describe periodic motion. Conversely, the exponential function $e^{-\omega t}$ continuously decays as time increases and never repeats its values, confirming it as non-periodic.

Hooke's Law establishes that the restoring force of a spring is directly proportional to the magnitude of its displacement from the equilibrium position. Because this force always acts in the opposite direction to the displacement, a negative sign is necessary to mathematically describe the relationship. This physical behavior is captured by the equation $F = -k x$, where $k$ represents the spring constant and $x$ stands for the displacement.

The displacement of a particle executing simple harmonic motion follows a sine function with a frequency of $n$, which is determined by the angular frequency $\omega$. Because the potential energy is proportional to the square of the displacement, it is expressed through a term involving $\sin^2(\omega t + \phi)$. Since the square of a sine wave oscillates at double the angular frequency of the original function, the potential energy completes two full cycles in the time it takes the particle to complete one cycle of motion. Consequently, the frequency of the potential energy is $2n$.

The fundamental relationship governing simple harmonic motion is that acceleration is directly proportional to the negative of displacement, represented by the equation $a = -\omega^2 y$. When we describe the position of a particle using the function $y = A \sin \omega t$, finding the acceleration requires taking the second derivative with respect to time, which results in $a = -A\omega^2 \sin \omega t$. By applying the trigonometric identity $-\sin \theta = \sin(\theta + \pi)$, we can rewrite the acceleration as $a = A\omega^2 \sin(\omega t + \pi)$, explicitly demonstrating that the phase difference between displacement and acceleration is $\pi$ radians.

Simple harmonic motion describes an oscillation where a particle swings between its two extreme points while passing through the equilibrium position. One complete time period corresponds to a full cycle of this motion. Starting from the equilibrium position, the particle travels to the positive extreme covering a distance $A$, returns to equilibrium covering $A$, moves to the negative extreme covering $A$, and finally returns to the equilibrium point adding another $A$. Summing these individual segments, the total distance covered in one time period is $4A$.

The amplitude of a simple harmonic oscillator represents the maximum displacement from its equilibrium position. Because the constant term $A_{0}$ merely shifts the mean position of the particle without affecting the oscillation magnitude, the oscillating portion of the motion is isolated by writing $y - A_{0} = A \sin \omega t + B \cos \omega t$. By treating the coefficients of the sine and cosine functions as orthogonal components, the resultant amplitude is determined using the Pythagorean theorem to be $\sqrt{A^{2} + B^{2}}$.

When a particle undergoes uniform circular motion, its projection onto the vertical axis oscillates in simple harmonic motion. The radius of the circular path serves as the amplitude of this motion, which is 3 m. Because the particle is at its maximum vertical displacement when $t = 0$, we describe the motion using a cosine function, $y(t) = A \cos(\omega t)$, since the cosine function is at its peak at time zero. Given the period of 4 s, the angular frequency is calculated as $\omega = \frac{2\pi}{4} = \frac{\pi}{2}$ rad/s, resulting in the displacement equation $y(t) = 3 \cos \left( \frac{\pi t}{2} \right)$.

Average velocity is defined as the total displacement divided by the total time interval taken. In simple harmonic motion, a particle completing one full vibration returns exactly to its starting point, which results in a net displacement of zero. Since the numerator in the formula $\text{Average velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}$ is zero, the resulting average velocity must also be zero.

Simple harmonic motion is fundamentally defined by a relationship where the magnitude of acceleration is directly proportional to the displacement from the equilibrium position, represented by $|a| = \omega^2 y$. Given an acceleration of 20 m/s² at a displacement of 5 m, we can calculate the angular frequency $\omega$ using this relationship.

$20 = \omega^2 \times 5$

Solving this equation leads to $\omega^2 = 4$, which means $\omega = 2$ rad/s. The time period of the oscillation is subsequently determined by $T = \frac{2\pi}{\omega}$, resulting in a time period of $\pi$ s.

The force constant of a spring is inversely proportional to its length, so dividing a spring into smaller segments increases their individual stiffness. Since the segments are cut in a ratio of $1 : 2 : 3$, the total length is effectively divided into six equal parts, resulting in pieces with lengths of $\frac{l}{6}$, $\frac{l}{3}$, and $\frac{l}{2}$. Applying the inverse proportionality rule, the new force constants for these segments are $k_1 = 6k$, $k_2 = 3k$, and $k_3 = 2k$.

In a series combination, the equivalent force constant $k'$ is determined by the equation:
$\frac{1}{k'} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3} = \frac{1}{6k} + \frac{1}{3k} + \frac{1}{2k} = \frac{1}{k}$
This calculation shows that $k' = k$. For the parallel combination, the equivalent force constant $k''$ is the sum of the individual constants:
$k'' = k_1 + k_2 + k_3 = 6k + 3k + 2k = 11k$
Comparing the series and parallel results establishes the ratio $k' : k'' = 1 : 11$.

In simple harmonic motion, the magnitude of the velocity at any displacement $x$ is defined by $|v| = \omega \sqrt{A^2 - x^2}$, while the magnitude of acceleration is given by $|a| = \omega^2 x$, with $A$ denoting amplitude and $\omega$ representing the angular frequency. Setting these two magnitudes equal requires that $\omega \sqrt{A^2 - x^2} = \omega^2 x$, which simplifies to the expression for frequency, $\omega = \frac{\sqrt{A^2 - x^2}}{x}$. Substituting the given values where $A = 3$ cm and $x = 2$ cm, we calculate $\omega = \frac{\sqrt{3^2 - 2^2}}{2} = \frac{\sqrt{5}}{2}$ radians per second. The time period is then found using the relation $T = \frac{2\pi}{\omega}$, which leads to $T = \frac{2\pi}{\frac{\sqrt{5}}{2}} = \frac{4\pi}{\sqrt{5}}$ seconds.

The time period of a mass oscillating on a spring is governed by $T = 2\pi \sqrt{\frac{m}{k}}$, showing that the period is proportional to the square root of the suspended mass. Since the spring constant $k$ remains consistent throughout the process, we can compare the two states by setting up the ratio $\frac{T_1}{T_2} = \sqrt{\frac{m_1}{m_2}}$. Substituting the provided values, we get $\frac{3}{5} = \sqrt{\frac{m}{m + 1}}$, and squaring both sides yields $\frac{9}{25} = \frac{m}{m + 1}$. Rearranging this equation leads to $9(m + 1) = 25m$, which expands to $9m + 9 = 25m$. Isolating the variable shows that $16m = 9$, which confirms that the initial mass is $m = \frac{9}{16}$ kg.

The velocity of a particle performing simple harmonic motion relates to its displacement $x$ from the mean position through the equation $V = \omega \sqrt{A^2 - x^2}$, where $A$ denotes the amplitude and $\omega$ represents the angular frequency. By squaring this relation for the two given states, we establish $V_1^2 = \omega^2(A^2 - x_1^2)$ and $V_2^2 = \omega^2(A^2 - x_2^2)$. Subtracting the second equation from the first eliminates the amplitude, resulting in $V_1^2 - V_2^2 = \omega^2(x_2^2 - x_1^2)$, which simplifies to $\omega = \sqrt{\frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}}$. Because the time period $T$ is related to angular frequency by $T = \frac{2\pi}{\omega}$, substituting the derived value of $\omega$ provides the expression $T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}$.

In simple harmonic motion, the maximum velocity of a particle is expressed as $\beta = A\omega$, while its maximum acceleration is given by $\alpha = A\omega^2$, where $A$ represents the amplitude and $\omega$ denotes the angular frequency. Dividing the expression for maximum acceleration by the expression for maximum velocity causes the amplitude to cancel out, leaving the ratio $\frac{\alpha}{\beta} = \omega$. Since the time period of vibration is defined as $T = \frac{2 \pi}{\omega}$, substituting the derived value for angular frequency yields the result $T = \frac{2 \pi \beta}{\alpha}$.

Acceleration in simple harmonic motion describes the second time derivative of the displacement. Calculating this requires two successive differentiations of the position function $X = A\cos(\omega t)$, where the first derivative provides the velocity $v = -A\omega\sin(\omega t)$, and the second results in the acceleration $a = -A\omega^2\cos(\omega t)$. This relationship indicates that the acceleration follows a negative cosine function over time, beginning at a negative maximum value of $-A\omega^2$ when $t=0$. The corresponding graph is a periodic wave that starts at this bottom peak and oscillates symmetrically above and below the horizontal time axis.

To determine the relationship between momentum and displacement for a particle in simple harmonic motion, we start with the position equation $x = a \sin \omega t$. Differentiating this expression with respect to time gives the velocity $v = a \omega \cos \omega t$, and since momentum is defined as the product of mass and velocity, it can be written as $p = m a \omega \cos \omega t$.

We can now express the trigonometric functions in terms of the variables $x$ and $p$ by rearranging these equations to get $\sin \omega t = \frac{x}{a}$ and $\cos \omega t = \frac{p}{m a \omega}$. Squaring both sides and applying the Pythagorean identity $\sin^2 \omega t + \cos^2 \omega t = 1$ leads to the following equation:
$\frac{x^2}{a^2} + \frac{p^2}{(m a \omega)^2} = 1$
This result is in the standard form of an ellipse, $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, where the semi-major and semi-minor axes are defined by the amplitude of displacement and the maximum momentum, respectively.

Simple harmonic motion can be analyzed by relating the instantaneous position of an oscillating particle to its phase angle. Given that both particles share the same frequency and amplitude, their states of motion at any specific displacement are determined by the time elapsed within their respective cycles. A particle traveling from the mean position at $x=0$ to $x=A/2$ requires a time interval of $T/12$, whereas a particle returning from the extreme position at $x=A$ reaches the same displacement of $x=A/2$ in a time of $T/6$.

Because the two particles pass each other at $x=A/2$ while traveling in opposite directions, the time difference between their states is the sum of these intervals. Combining these durations yields a total time difference of $T/6 + T/6 = T/3$. Converting this temporal offset into a phase difference using the angular frequency $\omega = \frac{2\pi}{T}$, the final phase difference is calculated as $\phi = \frac{2\pi}{T} \times \frac{T}{3} = \frac{2\pi}{3}$.