Oscillations – NEET UG PhysicsPractice Questions & PYQs

In simple harmonic motion, a particle achieves maximum kinetic energy at the mean equilibrium position, whereas its potential energy reaches a maximum at the extreme positions defined by the amplitude $a$. Because the equilibrium point corresponds to zero displacement and the extreme positions are located at $\pm a$, the displacement spanning from the location of maximum kinetic energy to the location of maximum potential energy is simply $\pm a$.

The oscillation of a mass suspended from a spring depends on its mass $m$ and the stiffness of the spring $k$, defined by the formula $T=2\pi\sqrt{m/k}$. For two separate springs with constants $k_1$ and $k_2$, the individual time periods are expressed as $t_1=2\pi\sqrt{m/k_1}$ and $t_2=2\pi\sqrt{m/k_2}$. When these springs are arranged in parallel, they share the load, creating an effective spring constant $k_{eff} = k_1 + k_2$, which results in a combined time period of $t_0 = 2\pi\sqrt{m/(k_1 + k_2)}$.

Squaring the individual time period equations allows us to isolate the spring constants, resulting in $1/t_1^2 = k_1/(4\pi^2m)$ and $1/t_2^2 = k_2/(4\pi^2m)$. Adding these two inverse squared terms provides $(k_1 + k_2)/(4\pi^2m)$, which is mathematically equivalent to $1/t_0^2$. This yields the relationship $t_0^{-2} = t_1^{-2} + t_2^{-2}$.

In simple harmonic motion, the total energy of an oscillating system is defined by the constant sum of its kinetic and potential energy. This total energy is expressed as $E = \frac{1}{2} k a^{2}$, which directly relates the energy to the force constant $k$ and the amplitude of the motion $a$. Since these variables fully account for the system's energy, the total energy remains independent of both mass and instantaneous displacement.

The oscillatory motion of a mass suspended from a spring depends on the spring constant, while the static displacement is determined by the balance between the weight of the object and the spring force. To find the spring constant $k$, we use the time period formula $T = 2 \pi \sqrt{\frac{m}{k}}$ By substituting the given period $T = 2 \pi$ and mass $m = 5$ into this expression, we have $2 \pi = 2 \pi \sqrt{\frac{5}{k}}$ which simplifies to $1 = \sqrt{\frac{5}{k}}$, establishing that $k = 5$. The change \in length when the mass is removed corresponds to the static extension, calculated using Hooke's Law as $\Delta x = \frac{mg}{k}$. Substituting the values leads to $\Delta x = \frac{5g}{5}$ which simplifies directly to $g$ metres.

In simple harmonic motion, the velocity $v$ of a particle at any displacement $x$ from the mean position is governed by the equation $v^2 = \omega^2 (A^2 - x^2)$, where $A$ represents the amplitude and $\omega$ is the angular frequency. By applying the two sets of provided data, we can establish a system of equations to determine the value of $\omega$. Substituting the first set, where $v = 10 \text{ cm/s}$ and $x = 4 \text{ cm}$, gives $100 = \omega^2 (A^2 - 16)$. Similarly, substituting the second set, where $v = 8 \text{ cm/s}$ and $x = 5 \text{ cm}$, gives $64 = \omega^2 (A^2 - 25)$.

To isolate the angular frequency, we express the equations as $\frac{100}{\omega^2} = A^2 - 16$ and $\frac{64}{\omega^2} = A^2 - 25$. Subtracting the second equation from the first eliminates the amplitude term, resulting in $\frac{100 - 64}{\omega^2} = (A^2 - 16) - (A^2 - 25)$. This simplifies to $\frac{36}{\omega^2} = 9$, which leads to $\omega^2 = 4$ and consequently $\omega = 2 \text{ rad/s}$. Since the time period $T$ is defined as $T = \frac{2 \pi}{\omega}$, substituting the calculated value of $\omega$ yields $T = \frac{2 \pi}{2} = \pi \text{ s}$.

Superimposing two perpendicular simple harmonic motions of equal amplitude and time period results in a trajectory defined by their relative phase difference. Representing the motion along the x-axis as $x = a \sin \omega t$ and along the y-axis as $y = a \sin(\omega t + \frac{\pi}{2})$, the y-expression simplifies to $y = a \cos \omega t$ due to the trigonometric phase shift. Squaring both equations and summing them yields $x^2 + y^2 = a^2(\sin^2 \omega t + \cos^2 \omega t)$, which simplifies to the relation $x^2 + y^2 = a^2$, identifying the path as a circle.

When a simple pendulum is released from an angular displacement $\theta$, the potential energy it possesses at the release point is entirely converted into kinetic energy as it reaches the equilibrium position. The vertical distance $h$ the bob descends is determined by subtracting the vertical projection of the string length at angle $\theta$ from the total length $l$, resulting in $h = l - l \cos\theta = l(1 - \cos\theta)$. By applying the principle of conservation of energy, where the potential energy $mgh$ equals the kinetic energy $\frac{1}{2}mv^2$, we simplify the expression to $v^2 = 2gh$ and substitute the value of $h$ to find the velocity $v = \sqrt{2gl(1 - \cos \theta)}$.

The oscillation period of a simple pendulum is determined by the formula $T = 2\pi \sqrt{\frac{l}{g}}$, which dictates that the time period decreases as the effective acceleration due to gravity increases. When a rocket moves upward with uniform acceleration, the pendulum experiences an increased effective gravity because of the added pseudo-force acting in the opposite direction of the motion. This resulting increase in the net gravitational pull causes the pendulum to oscillate more rapidly, thereby shortening its time period.

Any particle undergoing simple harmonic motion follows the fundamental relationship $a = \omega^2 x$, where the acceleration is defined by the product of the square of the angular frequency and the displacement from the equilibrium position. By rearranging this physical law to solve for the angular frequency, we obtain the following expression:

$\omega = \sqrt{\frac{a}{x}}$

Given that the acceleration is 2.0 and the displacement is 0.02, we substitute these values into the formula. Dividing 2.0 by 0.02 yields 100, and calculating the square root of 100 provides an angular frequency of 10 rad s-1.

The time period of a simple pendulum depends on the effective acceleration acting on the bob, which is the vector resultant of all individual accelerations present in the non-inertial frame. Because the trolley is accelerating horizontally with $a$ and gravity acts vertically with $g$, the bob experiences these two mutually perpendicular components, leading to an effective gravitational acceleration magnitude of $g_{eff} = \sqrt{g^2 + a^2}$. Substituting this resultant value into the denominator of the standard time period formula provides the physically accurate time period for the accelerating system.

Simple harmonic motion dictates that the maximum speed of an oscillator occurs at the mean position and is given by $v_{max} = A \omega$, while the speed at any arbitrary displacement $y$ is expressed as $v = \omega \sqrt{A^2 - y^2}$. Setting the particle's speed to half of its maximum value leads to the relationship
$\frac{A \omega}{2} = \omega \sqrt{A^2 - y^2}$
Canceling the angular frequency $\omega$ from both sides and squaring the remaining terms results in
$\frac{A^2}{4} = A^2 - y^2$
Rearranging this expression to isolate $y^2$ gives $y^2 = A^2 - \frac{A^2}{4} = \frac{3A^2}{4}$, which simplifies to a final displacement of $y = \frac{\sqrt{3} A}{2}$.

The frequency of a simple pendulum undergoing simple harmonic motion is determined solely by the length of the string and the acceleration due to gravity, as defined by the formula $f = \frac{1}{2 \pi} \sqrt{\frac{g}{l}}$. Notably, this frequency is entirely independent of the mass of the bob, meaning any variation in the mass does not influence the oscillation rate. Given that the frequency of pendulum A is twice that of pendulum B, we set up the equation $\frac{1}{2 \pi} \sqrt{\frac{g}{l_{A}}} = 2 \times \frac{1}{2 \pi} \sqrt{\frac{g}{l_{B}}}$. Simplifying this by removing the common factors of $2 \pi$ and $\sqrt{g}$ from both sides results in $\frac{1}{\sqrt{l_{A}}} = \frac{2}{\sqrt{l_{B}}}$. Squaring both sides of this expression yields $\frac{1}{l_{A}} = \frac{4}{l_{B}}$, which confirms that $l_{A} = \frac{l_{B}}{4}$.

The oscillation frequency of a simple pendulum is determined by the relation $f = \frac{1}{2 \pi} \sqrt{\frac{g}{L}}$, indicating that frequency is inversely proportional to the square root of the pendulum's length. By comparing the two pendulums, the ratio of the frequency of the $5 \text{ m}$ pendulum to that of the $20 \text{ m}$ pendulum is calculated as $\frac{f_{5}}{f_{20}} = \sqrt{\frac{20}{5}}$, which simplifies to $2$. This means the shorter pendulum oscillates twice as fast as the longer one, completing $2$ oscillations in the exact time it takes the longer pendulum to complete $1$ oscillation, at which point they will return to the same phase.

The time period of a simple pendulum is governed by the formula $T = 2 \pi \sqrt{\frac{l}{g}}$, which establishes that the time period is directly proportional to the square root of the length, or $T \propto \sqrt{l}$. When dealing with small percentage changes in a physical quantity, the fractional change in the dependent variable is determined by multiplying the exponent of the independent variable by its fractional change. Since the length in this relationship is raised to the power of $1/2$, the percentage change in the time period is half of the percentage change in length. Applying this to a $2\%$ increase in length leads to an increase of $1/2 \times 2\%$, which results in a $1\%$ increase in the time period.

The oscillation of a simple pendulum is governed by the conservation of mechanical energy, where the gravitational potential energy at the extreme points $A$ or $C$ converts completely into kinetic energy at the lowest point $B$. Since the bob is at a vertical height $H$ relative to point $B$, its initial potential energy is $m g H$. As the bob swings through $B$, this energy transforms entirely into kinetic energy expressed as $\frac{1}{2} m v_B^2$. By setting these two quantities equal, the mass $m$ cancels out, leading to the relationship $\frac{1}{2} v_B^2 = g H$, which solves to $v_B = \sqrt{2 g H}$.

The behavior of a simple pendulum is governed by the equation $T = 2\pi\sqrt{\frac{l}{g}}$, which demonstrates that the time period is directly proportional to the square root of the length, or $T \propto \sqrt{l}$. When the length is multiplied by 4, the time period changes by a factor of $\sqrt{4}$, which simplifies to 2. Applying this factor to the original duration of 2 seconds results in a new time period of 4 seconds.

When two springs are connected in series, the system acts as a single equivalent spring characterized by an effective force constant $k$ derived from the harmonic addition of the individual constants:
$k = \frac{k_1 k_2}{k_1 + k_2}$
This relationship arises because the total extension of the system under an applied force is the sum of the individual extensions of each spring.

The time period for a mass oscillating in simple harmonic motion is given by the standard formula $T = 2 \pi \sqrt{\frac{m}{k}}$. Substituting the expression for the equivalent spring constant into this equation involves dividing by the fraction, which results in:
$T = 2 \pi \sqrt{\frac{m (k_1 + k_2)}{k_1 k_2}}$
This final expression defines the period of oscillation for the mass $m$ attached to the series-connected spring arrangement.

The frequency of an oscillating spring-mass system is inversely proportional to the square root of the suspended mass, which is described by the equation
$n = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$
where $k$ represents the spring constant. When the mass is increased to $4m$, the expression for the new frequency $n'$ becomes
$n' = \frac{1}{2 \pi} \sqrt{\frac{k}{4m}}$
Because the square root of $4$ in the denominator evaluates to $2$, this simplifies to $n' = n / 2$, indicating that the frequency of the system is reduced to half its original value.