Oscillations – NEET UG PhysicsPractice Questions & PYQs

A motion qualifies as Simple Harmonic Motion when the displacement is described by a single sinusoidal function, $y = A \sin(\omega t + \phi)$ or $y = A \cos(\omega t + \phi)$. The expression $y = \sin \omega t - \cos \omega t$ can be converted into a single harmonic form, which represents Simple Harmonic Motion with a time period of $T = \frac{2\pi}{\omega}$. The function $y = \sin^3 \omega t$ deviates from this definition because the identity $\sin^3 \theta = \frac{3 \sin \theta - \sin 3\theta}{4}$ reveals that it consists of two distinct harmonic terms with different frequencies, resulting in motion that is periodic but not simple harmonic.

Regarding $y = 5 \cos(\frac{3\pi}{2} - 3\omega t)$, this is a standard sinusoidal function that represents Simple Harmonic Motion with a time period of $T = \frac{2\pi}{3\omega}$. Lastly, $y = 1 + \omega t + \omega^2 t^2$ represents a quadratic polynomial in time that lacks the bounded, oscillatory nature required for periodic motion, confirming that it does not represent Simple Harmonic Motion.

Simple harmonic motion is defined by the requirement that the restoring acceleration of a particle is directly proportional to its displacement from the origin, satisfying the condition $a \propto -x$. To evaluate the nature of the motion given by $x = \sin^2\omega t$, we apply the trigonometric identity $\sin^2\theta = \frac{1-\cos 2\theta}{2}$ to rewrite the displacement as $x = \frac{1}{2} - \frac{1}{2}\cos(2\omega t)$. Differentiating this expression twice with respect to time to find the acceleration gives $a = \frac{d^2x}{dt^2} = 2\omega^2 \cos(2\omega t)$. Because the acceleration is proportional to the cosine function rather than the negative of the displacement function $x$, the motion fails to meet the fundamental requirement for simple harmonic motion, characterizing it as non-simple harmonic.

The period of oscillation for a spring-mass system is determined by the relationship $T = 2\pi\sqrt{\frac{M}{k}}$, where $M$ is the suspended mass and $k$ is the spring constant. When an additional mass $M$ is attached, the total mass of the system effectively doubles to $2M$, causing the new period to become $T' = 2\pi\sqrt{\frac{2M}{k}}$. By isolating the constant $\sqrt{2}$ from the expression, this simplifies to $T' = \sqrt{2} \times (2\pi\sqrt{\frac{M}{k}})$, confirming that the new oscillation period is $\sqrt{2}T$.

Simple harmonic motion is defined by an acceleration that is directly proportional to the displacement from the equilibrium position and is always directed towards that position, expressed mathematically as $a = -\omega^2 x'$, where $x'$ is the displacement from equilibrium. The equation $a = -k(x+a)$ represents motion about an equilibrium position shifted to $x = -a$, where the term $(x+a)$ acts as the displacement variable and the negative sign confirms the restoring nature of the acceleration. Because this equation maintains the required linear relationship between acceleration and displacement, it satisfies the fundamental criteria for simple harmonic motion.

The velocity of a simple harmonic oscillator depends directly on its displacement from the mean position according to the relationship $v = \omega \sqrt{a^2 - x^2}$, where $a$ defines the amplitude and $\omega$ represents angular frequency. Substituting the specific displacement $x = \frac{a}{2}$ into the velocity equation leads to $v = \omega \sqrt{a^2 - \frac{a^2}{4}}$, which simplifies to $v = \omega \sqrt{\frac{3a^2}{4}}$ or $v = \frac{\omega a \sqrt{3}}{2}$. Introducing the relationship between angular frequency and the time period, where $\omega = \frac{2\pi}{T}$, allows the velocity to be expressed as $v = \frac{2\pi}{T} \cdot \frac{a \sqrt{3}}{2}$, resulting in $v = \frac{\pi a \sqrt{3}}{T}$.

In simple harmonic motion, the maximum acceleration is defined by the formula $a_{\mathrm{max}} = \omega^2 A$, where $\omega$ is the angular frequency and $A$ is the displacement amplitude. Because the amplitude remains the same for both motions, the ratio of their maximum accelerations is determined entirely by the ratio of the squares of their angular frequencies:

$\frac{a_{\mathrm{max}_1}}{a_{\mathrm{max}_2}} = \frac{\omega_1^2}{\omega_2^2}$

Substituting the given frequencies of 100 rads^-1 and 1000 rads^-1 into this expression leads to $\frac{100^2}{1000^2}$, which simplifies to $\frac{1}{100}$. Therefore, the resulting ratio of the maximum accelerations is $1:10^2$.

In simple harmonic motion, the displacement of a particle is described by the function $y = A \sin(\omega t)$. Since velocity is the time derivative of displacement, it incorporates a phase shift of $\frac{\pi}{2}$, resulting in $v = A \omega \sin(\omega t + \frac{\pi}{2})$, while acceleration is the further derivative of velocity, which introduces another phase shift to reach $a = A \omega^2 \sin(\omega t + \pi)$. Subtracting the phase of velocity from the phase of acceleration confirms a constant phase difference of $0.5 \pi$.

The mass loses contact with the pan when the downward acceleration of the pan during its simple harmonic motion equals the acceleration due to gravity, $g$. In a vertical spring-mass system, the equilibrium position is established where the spring compression balances the weight of the mass, resulting in a compression of $x_0 = \frac{mg}{k}$. At this point, the spring exerts an upward force of $mg$ to support the mass.

For the pan to detach, its downward acceleration at the highest point of the oscillation must exceed the gravitational acceleration. The acceleration of the pan in simple harmonic motion is given by $a = \omega^2 x$, where $x$ is the displacement from equilibrium and $\omega^2 = \frac{k}{m}$. Setting the acceleration equal to $g$ at the extreme displacement $A$, the condition for detachment becomes $g = \frac{k}{m} A$. Solving for the amplitude yields $A = \frac{mg}{k}$. Using the values $m$ = 2.0 kg, $g$ = 10 m/s^2, and $k$ = 200 N/m, the minimum amplitude is 0.10 m, which is 10.0 cm.

The displacement of a particle executing simple harmonic motion starting from its equilibrium position is given by the equation $x(t) = a \sin(\omega t)$, where $\omega = \frac{2\pi}{T}$. To find the time taken to reach a displacement of $\frac{a}{2}$, we substitute this value into the equation as follows:
$\frac{a}{2} = a \sin\left(\frac{2\pi t}{T}\right)$
Dividing both sides by $a$ yields $\sin\left(\frac{2\pi t}{T}\right) = \frac{1}{2}$. Because $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, we equate the arguments $\frac{2\pi t}{T} = \frac{\pi}{6}$, which simplifies to $t = \frac{T}{12}$.

In simple harmonic motion, the total energy is the sum of kinetic and potential energies and remains constant throughout the oscillation. The maximum kinetic energy occurs when the potential energy is zero, and conversely, the maximum potential energy occurs when the kinetic energy is zero, meaning both are equal to the total energy. Given the kinetic energy expression $K = K_0 \cos^2 \omega t$, the maximum kinetic energy is $K_0$, which must therefore equal the total energy. Consequently, the maximum potential energy and the total energy are both $K_0$.

When a block of mass $m$ floats in equilibrium, its weight is balanced by the buoyant force. Applying a small vertical displacement $y$ increases the submerged volume by $Ay$, resulting in an additional upward buoyant force equal to $Ay \rho g$. Since this restoring force $F = -(A \rho g)y$ follows the form of Hooke's law $F = -ky$, the system acts as a simple harmonic oscillator with an effective spring constant $k = A \rho g$. Substituting these parameters into the standard period formula $T = 2 \pi \sqrt{\frac{m}{k}}$, we find that $T = 2 \pi \sqrt{\frac{m}{A \rho g}}$, which clearly demonstrates that $T$ is inversely proportional to the square root of the cross-sectional area $A$.

Periodic motion refers to any movement that repeats its cycle over a consistent time interval, and since a particle traveling around a circle at a steady pace returns to its starting position at regular intervals, it qualifies as periodic. Simple Harmonic Motion requires that the acceleration of the particle be directly proportional to its displacement from a mean position, typically governed by the relationship $a \propto -x$. Because the centripetal acceleration in uniform circular motion remains constant in magnitude at $a_c = \frac{v^2}{r}$ rather than varying linearly with displacement, the motion lacks the specific restoring force dynamic necessary to be classified as simple harmonic.

When springs are connected in series, a single applied force $F$ acts on the entire combination, causing each spring to stretch by an amount dependent on its stiffness such that the total extension $y$ is simply the sum of individual displacements, $y = y_1 + y_2$. Substituting the standard displacement formula $y = F/k$ into this sum gives the equation $F/k_{eff} = F/k_1 + F/k_2$. By dividing both sides by the constant force $F$, we arrive at the reciprocal addition rule for springs in series, which simplifies to $k_{eff} = \frac{k_1 k_2}{k_1 + k_2}$.

The maximum speed of a particle undergoing simple harmonic motion is given by the product of its angular frequency and amplitude, defined as $v_{\text{max}} = \omega A$. Since angular frequency relates to ordinary frequency through the expression $\omega = 2\pi f$, we can rewrite the governing equation as $v_{\text{max}} = 2\pi f A$. Substituting the given values of $31.4 \text{ cm/s}$ for the maximum speed and $5 \text{ cm}$ for the amplitude, the equation becomes $31.4 = 2 \times 3.14 \times f \times 5$. Solving this further, $31.4 = 10 \times 3.14 \times f$, which simplifies to a frequency of $1 \text{ Hz}$.

The total energy $E$ of a simple harmonic oscillator is proportional to the square of its amplitude, given by the expression $E = \frac{1}{2} m \omega^2 a^2$. The potential energy of the oscillator at any displacement $y$ is determined by the formula $U = \frac{1}{2} m \omega^2 y^2$. When the particle is situated at half its end point, its displacement is $y = \frac{a}{2}$. Substituting this value into the potential energy equation gives $U = \frac{1}{2} m \omega^2 (\frac{a}{2})^2$, which simplifies to $U = \frac{1}{4} (\frac{1}{2} m \omega^2 a^2)$. Since the expression in the parenthesis represents the total energy of the system, the potential energy at this position is $\frac{1}{4} E$.

The stiffness of a spring, represented by its force constant $k$, is inversely proportional to its length, which means that cutting the spring into four equal segments increases the stiffness of each part to $k' = 4k$. When the mass is suspended from one of these new, stiffer segments, the expression for the new time period is derived by substituting $k'$ into the fundamental formula $T = 2\pi\sqrt{m/k}$, resulting in:
$T' = 2\pi\sqrt{\frac{m}{4k}}$
Simplifying this by factoring the constant out of the square root yields:
$T' = \frac{1}{2} \times 2\pi\sqrt{\frac{m}{k}} = \frac{T}{2}$
This calculation confirms that the new time period is exactly half of the original value.

A particle undergoing simple harmonic motion experiences a restoring force directly proportional to its displacement, which results in a potential energy function defined by the quadratic relationship $U = \frac{1}{2}kx^2$. This mathematical dependence indicates that the energy profile takes the shape of an upward-opening parabola, touching the horizontal axis at the equilibrium point where displacement is zero and reaching its peak values at the extreme displacement positions. Consequently, the visual representation of potential energy versus position must be a parabolic curve that is minimal at the mean position and rises towards the boundaries of the motion.

In simple harmonic motion, the particle's acceleration is governed by the relationship $a = -\omega^2 x$, which demonstrates that acceleration is directly proportional to displacement but acts in the opposite direction. At the equilibrium position, where the displacement $x$ is zero, the particle reaches its peak velocity $v = A\omega$, and consequently, the acceleration must be zero. Conversely, at the extreme positions where the velocity momentarily drops to zero, the displacement is at its greatest magnitude, which results in the maximum value for acceleration.

The sharpness of a resonance peak in forced vibration is primarily governed by the level of damping present in the system, which determines the rate of energy dissipation. When the damping force is small, the system loses minimal energy per cycle, allowing the amplitude of oscillation to increase significantly as the driving frequency aligns with the natural frequency. This physical condition results in a taller and narrower resonance peak, creating a more pronounced and sharply defined response profile.

In systems undergoing damped oscillation, the amplitude decreases exponentially over time according to the relation $A = A_0 e^{-bt}$. Because the number of oscillations completed is directly proportional to the elapsed time, we can model the amplitude after $n$ cycles using the factor $e^{-bT}$, where $T$ is the time period. Given that 100 oscillations result in the amplitude becoming $\frac{1}{3}$ of the initial value, we have the following relationship:
$e^{-100bT} = \frac{1}{3}$
When the oscillator completes 200 oscillations, the amplitude follows the expression $A = A_0 e^{-200bT}$. By recognizing that $e^{-200bT}$ is equivalent to the square of the decay factor observed at 100 oscillations, we can determine the new amplitude as:
$A = A_0 \left(e^{-100bT}\right)^2 = A_0 \left(\frac{1}{3}\right)^2 = \frac{A_0}{9}$
The amplitude is reduced to $\frac{1}{9}$ of its initial value.