Q.A particle is in linear simple harmonic motion between two extreme points A and B lying on a straight line $10\ \text{cm}$ apart. The mean (equilibrium) position O is the midpoint, so $AO=OB=5\ \text{cm}$. A point C lies between O and A such that $BC=8\ \text{cm}$; hence C is $3\ \text{cm}$ from O on the A-side ($OC=3\ \text{cm}$, and C is $2\ \text{cm}$ from A). Reading along the line, the order of the points is B, O, C, A. Take the direction from A to B as the positive direction, and choose the correct statement(s). (Note: more than one of the given options may be correct.)

Showing the 12 most recent of 14 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.Force constant of interatomic bond, in a certain element, is $7.1 \mathrm{Nm}^{-1}$. If the atom oscillates in SHM in a certain direction, what is its frequency? Given: Mole weight of the given element is 108 g and Avagadro's number $=6.023 \times 10^{23} \mathrm{~g} \mathrm{~mol}^{-1}$ (A) $3.45 \times 10^{22} s^{-1}$ (B) $0.005 \times 10^{12} s^{-1}$ (C) $1 \times 10^{12} s^{-1}$ (D) $6.667 \times 10^{12} s^{-1}$

   ›Reveal solutionSolution

   The frequency of an atom oscillating in SHM is found from the spring constant and the mass of a single atom. Using the force constant $k = 7.1\ \mathrm{N/m}$ and the atomic mass (from mole weight and Avogadro’s number), the frequency is $1 \times 10^{12}\ \mathrm{s^{-1}}$, which corresponds to option (C).

   ---

   Concept & Intuition

   An atom in a crystal bonded to its neighbours can be modelled as a simple harmonic oscillator: the interatomic bond acts like a tiny spring with force constant $k$. The frequency of oscillation for a mass $m$ on a spring is $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$. Here the “mass” is the mass of a single atom. We are given the molar mass (108 g/mol) and Avogadro’s number, so we can find the mass of one atom. Then plug into the formula.

   ---

   Step-by-step solution

   1. Find the mass of a single atom Molar mass $M = 108\ \mathrm{g/mol} = 0.108\ \mathrm{kg/mol}$ (convert to kg for SI consistency). Avogadro’s number $N_A = 6.023 \times 10^{23}\ \mathrm{atoms/mol}$. Mass of one atom:

   $m = \frac{M}{N_A} = \frac{0.108}{6.023 \times 10^{23}} \ \mathrm{kg}$

   Compute:

   $m = 1.793 \times 10^{-25}\ \mathrm{kg}$

   2. Apply the SHM frequency formula For a mass $m$ on a spring of constant $k$, the angular frequency $\omega = \sqrt{k/m}$, so the ordinary frequency is

   $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

   Given $k = 7.1\ \mathrm{N/m}$,

   $f = \frac{1}{2\pi} \sqrt{\frac{7.1}{1.793 \times 10^{-25}}}$

   3. Simplify inside the square root

   $\frac{7.1}{1.793 \times 10^{-25}} = \frac{7.1}{1.793} \times 10^{25} \approx 3.96 \times 10^{25}$

   (More precisely: $7.1 / 1.793 \approx 3.960$)

   4. Take the square root $\sqrt{3.96 \times 10^{25}} = \sqrt{3.96} \times 10^{12.5} \approx 1.99 \times 10^{12.5}$ …

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2. COMEDK 2025Set 2025-A1 markMCQ
   Q.Time period of oscillation of a mass suspended from a spring is T . If the spring is cut into four equal parts and the same mass is suspended from one of the parts, the fractional change in time period is (A) $\frac{1}{8}$ (B) $\frac{1}{4}$ (C) $\frac{1}{2}$ (D) $\frac{1}{6}$

   ›Reveal solutionSolution

   Cutting a spring into four equal parts increases its spring constant by a factor of 4, which reduces the time period by half, giving a fractional change of $1/2$.

   Concept & Intuition

   The time period of a mass–spring system depends only on the mass and the spring constant: $T = 2\pi \sqrt{m/k}$. When you cut a uniform spring into equal pieces, each piece becomes stiffer — the spring constant is inversely proportional to the length. So cutting into four parts makes each part’s constant four times larger. Since $T \propto 1/\sqrt{k}$, a fourfold increase in $k$ halves the period. The fractional change is then $(T_{\text{new}} - T_{\text{original}})/T_{\text{original}}$.

   Step-by-step reasoning

   1. Original time period For a spring of constant $k$ and mass $m$:

   $T = 2\pi \sqrt{\frac{m}{k}}.$

   2. Effect of cutting the spring A spring’s constant is inversely proportional to its length: $k \propto 1/L$. Cutting into four equal parts means each part has length $L/4$, so its spring constant becomes

   $k' = 4k.$

   3. New time period With the same mass $m$ suspended from one part:

   $T' = 2\pi \sqrt{\frac{m}{k'}} = 2\pi \sqrt{\frac{m}{4k}} = \frac{1}{2} \cdot 2\pi \sqrt{\frac{m}{k}} = \frac{T}{2}.$

   4. Fractional change Fractional change is defined as $$\frac{T - T'}{T} = \frac{T - T/2}{T} = \frac{T/2}{T} = \frac{1}{2}. …$$

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3. COMEDK 2025Set 2025-M1 markMCQ
   Q.Two simple harmonic motions are represented by equations $y_1=0.5 \sin \left[200 \pi \mathrm{t}+\frac{\pi}{3}\right]$ and $y_1=0.5 \cos \pi \mathrm{t}$. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is : (A) $\frac{\pi}{2}$ (B) $\frac{\pi}{6}$ (C) $\frac{-\pi}{6}$ (D) $\frac{-\pi}{2}$

   ›Reveal solutionSolution

   The phase difference between the velocities is found by differentiating each displacement, writing both velocities as sine functions, and subtracting their phases. The result is $-\pi/6$, so option (C) is correct.

   Concept & Intuition

   The phase difference between two oscillating quantities is the difference in their arguments when both are expressed using the same trigonometric function (e.g., both as sine or both as cosine). Here we have two SHM displacements, but the question asks for the phase difference of their velocities. Velocity is the time derivative of displacement, which shifts the phase by $+\pi/2$ for a sine (since $\frac{d}{dt}\sin(\omega t+\phi) = \omega\cos(\omega t+\phi) = \omega\sin(\omega t+\phi+\pi/2)$). So we differentiate each, rewrite both as sine functions, then subtract phases.

   Step-by-step solution

   1. Write the given displacements clearly

   $y_1 = 0.5 \sin\!\left(200\pi t + \frac{\pi}{3}\right), \qquad y_2 = 0.5 \cos(\pi t).$

   2. Find the velocity of particle 1 Differentiate $y_1$ with respect to time:

   $v_1 = \frac{dy_1}{dt} = 0.5 \cdot 200\pi \cos\!\left(200\pi t + \frac{\pi}{3}\right) = 100\pi \cos\!\left(200\pi t + \frac{\pi}{3}\right).$

   Convert cosine to sine using $\cos\theta = \sin(\theta + \pi/2)$:

   $v_1 = 100\pi \sin\!\left(200\pi t + \frac{\pi}{3} + \frac{\pi}{2}\right) = 100\pi \sin\!\left(200\pi t + \frac{5\pi}{6}\right).$

   3. Find the velocity of particle 2 Differentiate $y_2$:

   $v_2 = \frac{dy_2}{dt} = 0.5 \cdot (-\pi) \sin(\pi t) = -0.5\pi \sin(\pi t).$

   Write the negative sign as a phase shift: $-\sin\theta = \sin(\theta + \pi)$ (since $\sin(\theta+\pi)=-\sin\theta$). So

   $v_2 = 0.5\pi \sin(\pi t + \pi).$

   Alternatively, we could also write $v_2 = 0.5\pi \sin(\pi t - \pi)$; either is fine — the key is to have a consistent sine form.

   4. Determine the phase difference The phase of $v_1$ is $\phi_1 = 200\pi t + \frac{5\pi}{6}$. The phase of $v_2$ is $\phi_2 = \pi t + \pi$. The phase difference (velocity of 1 with respect to velocity of 2) is $\Delta\phi = \phi_1 - \phi_2 = \left(200\pi t + \frac{5\pi}{6}\right) - \left(\pi t + \pi\right) = (200\pi - \pi)t + \frac{5\pi}{6} - \pi.$ …

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4. COMEDK 2025Set 2025-M1 markMCQ
   Q.A bar magnet is oscillating in the earth's magnetic field with a period $T$. When the length of the bar magnet is doubled and its mass is quadrupled, the time period is $T_1$. The ratio of $T_1$ to $T$ is (A) $1: 2$ (B) $4: 1$ (C) $2: 1$ (D) $1: 4$

   ›Reveal solutionSolution

   With $T \propto \sqrt{I}$ (magnetic moment fixed) and $I=ML^2/12$, quadrupling mass and doubling length multiplies $I$ by $16$, so $T_1/T = \sqrt{16} = 4$, i.e. $4:1$.

   The time period of a bar magnet oscillating in field $B$:

   $T = 2\pi\sqrt{\frac{I}{m B}},$

   where $I$ is the moment of inertia and $m$ the magnetic moment.

   Moment of inertia of a bar about its centre: $I = \dfrac{ML^2}{12}$.

   New values: mass $\to 4M$, length $\to 2L$: …

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5. COMEDK 2024Set 2024-A1 markMCQ
   Q.A circular disc of mass $20 \mathrm{~kg}$, having radius $10 \mathrm{~cm}$ is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The time period of torsional oscillations is found to be $1 \mathrm{~s}$. The torsional spring constant of the wire is (A) $6 \mathrm{~N~m} \mathrm{~rad}^{-1}$ (B) $9.86 \mathrm{~N~m} \mathrm{~rad}^{-1}$ (C) $3.94 \mathrm{~N~m} \mathrm{~rad}^{-1}$ (D) $1.264 \mathrm{~N~m} \mathrm{~rad}^{-1}$

   ›Reveal solutionSolution

   The disc's moment of inertia is $0.1$ kg·m²; a torsional period of $1$ s gives spring constant $k=4\pi^2 I\approx3.94$ N·m/rad.

   Moment of inertia of a disc about its central axis:

   $I=\frac{1}{2}MR^{2}=\frac{1}{2}(20)(0.10)^{2}=0.1\ \text{kg m}^{2}.$

   Torsional period: $T=2\pi\sqrt{\dfrac{I}{k}}$, so …

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6. COMEDK 2024Set 2024-E1 markMCQ
   Q.A body is executing SHM. When its displacements from the mean position are $4 \mathrm{~cm}$ and $5 \mathrm{~cm}$ it has velocity $10 \mathrm{~cms}^{-1}$ and $8 \mathrm{~cms}^{-1}$ respectively. Its periodic time $\mathrm{t}$ is (A) $\frac{2 \pi}{3} \sec$ (B) 2$\pi$ sec (C) $\frac{3 \pi}{2} \mathrm{sec}$ (D) $\pi$ sec

   ›Reveal solutionSolution

   Using the SHM velocity-displacement relation $v^2 = \omega^2 (A^2 - x^2)$ with two data pairs, we solve for $\omega$ and then period $T = 2\pi/\omega$. The period comes out to $\pi$ seconds, so option (D) is correct.

   Concept & Intuition

   In simple harmonic motion, the velocity is not constant — it depends on where the particle is. The relation $v^2 = \omega^2 (A^2 - x^2)$ ties velocity $v$, displacement from mean $x$, angular frequency $\omega$, and amplitude $A$. If we have two different $(x, v)$ pairs, we can set up two equations and eliminate $A$ to find $\omega$ directly. Once we have $\omega$, the period is $T = 2\pi/\omega$.

   1. Write the SHM velocity equation for each data point. For a particle in SHM:

   $v^2 = \omega^2 (A^2 - x^2)$

   Given:

   • When $x_1 = 4\ \text{cm}$, $v_1 = 10\ \text{cm/s}$
   • When $x_2 = 5\ \text{cm}$, $v_2 = 8\ \text{cm/s}$

   So we have:

   $10^2 = \omega^2 (A^2 - 4^2) \quad\Rightarrow\quad 100 = \omega^2 (A^2 - 16) \qquad(1)$

   $8^2 = \omega^2 (A^2 - 5^2) \quad\Rightarrow\quad 64 = \omega^2 (A^2 - 25) \qquad(2)$

   2. Eliminate $A^2$ by subtracting the equations. Subtract (2) from (1):

   $100 - 64 = \omega^2 (A^2 - 16) - \omega^2 (A^2 - 25)$

   $36 = \omega^2 \big[(A^2 - 16) - (A^2 - 25)\big] = \omega^2 (9)$

   So:

   $$\omega^2 = \frac{36}{9} = 4 \quad\Rightarrow\quad \omega = 2\ \text{rad/s} …$$

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7. COMEDK 2024Set 2024-M1 markMCQ
   Q.Two charges '$-q$' each are fixed, separated by distance '$2 d$'. A third charge '$q$' of mass '$m$' placed at the mid-point is displaced slightly by '$x$' $(x< < d)$ perpendicular to the line joining the two fixed charges as shown in Fig. The time period of oscillation of '$q$' will be (A) $\mathrm{T}=\sqrt{\frac{8 \varepsilon_0 \mathrm{~m} \pi^2 \mathrm{~d}^3}{\mathrm{q}^2}}$ (B) $\mathrm{T}=\sqrt{\frac{8 \varepsilon_0 \mathrm{~m} \pi^3 \mathrm{~d}^3}{\mathrm{q}^3}}$ (C) $\mathrm{T}=\sqrt{\frac{4 \varepsilon_0 \mathrm{~m} \pi^3 \mathrm{~d}^3}{\mathrm{q}^2}}$ (D) $\mathrm{T}=\sqrt{\frac{8 \varepsilon_0 \mathrm{~m} \pi^3 \mathrm{~d}^3}{\mathrm{q}^2}}$

   ›Reveal solutionSolution

   The problem reduces to simple harmonic motion because, for small perpendicular displacement, the net restoring force is proportional to the displacement. The time period is $T = 2\pi \sqrt{\frac{m}{k}}$, leading to $T = \sqrt{\frac{8\varepsilon_0 m \pi^3 d^3}{q^2}}$, which matches option (D).

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   Concept & Intuition

   When the third charge $q$ is exactly at the midpoint, the forces from the two fixed $-q$ charges are equal and opposite, so the net force is zero. If we displace it slightly perpendicular to the line joining the fixed charges, symmetry ensures the horizontal components of the forces cancel, but the vertical components add. For small $x \ll d$, the net vertical force is approximately linear in $x$ — that’s the hallmark of simple harmonic motion. So we can find an effective spring constant $k$ and then the period $T = 2\pi\sqrt{m/k}$.

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   Step-by-step solution

   1. Set up the geometry and forces The two fixed charges are at $(-d,0)$ and $(d,0)$. The movable charge $q$ is at $(0,x)$ with $x \ll d$. Distance from the movable charge to each fixed charge:

   $r = \sqrt{d^2 + x^2}.$

   The magnitude of the force from one fixed charge ($-q$) on the movable charge ($q$) is given by Coulomb’s law:

   $F = \frac{1}{4\pi\varepsilon_0} \cdot \frac{|q \cdot (-q)|}{r^2} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q^2}{d^2 + x^2}.$

   Since the charges are opposite, the force is attractive — each fixed charge pulls the movable charge toward itself.

   2. Resolve forces into components By symmetry, the horizontal components from the two fixed charges cancel. The vertical components add. From the geometry (see the figure), the angle $\theta$ satisfies $\sin\theta = \frac{x}{r}$. The vertical component of the force from one fixed charge is

   $F_y = F \sin\theta = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q^2}{d^2 + x^2} \cdot \frac{x}{\sqrt{d^2 + x^2}}.$

   So the net vertical force (both charges pulling downward, opposite to the displacement direction if we take upward as positive) is

   $F_{\text{net}} = -2 \cdot \frac{1}{4\pi\varepsilon_0} \cdot \frac{q^2 x}{(d^2 + x^2)^{3/2}}.$

   The negative sign indicates the force is restoring (toward the midpoint).

   3. Apply the small‑displacement approximation For $x \ll d$, we expand $(d^2 + x^2)^{-3/2}$ using the binomial approximation:

   $(d^2 + x^2)^{-3/2} = d^{-3} \left(1 + \frac{x^2}{d^2}\right)^{-3/2} \approx d^{-3} \left(1 - \frac{3}{2}\frac{x^2}{d^2} + \cdots\right).$

   Keeping only the leading term (since $x^2/d^2$ is negligible), we get

   $$$$

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8. KCET 2023Set A-31 markMCQ
   Q.In series LCR circuit at resonance, the phase difference between voltage and current is (A) $\pi$ (B) $\frac{\pi}{4}$ (C) $\frac{\pi}{2}$ (D) zero

   ›Reveal solutionSolution

   At resonance in a series LCR circuit, the inductive and capacitive reactances cancel, making the circuit purely resistive — so voltage and current are in phase, meaning the phase difference is zero.

   The key idea is that resonance in a series LCR circuit is defined by the condition where the inductive reactance $X_L = \omega L$ exactly equals the capacitive reactance $X_C = \frac{1}{\omega C}$. When that happens, the net reactance $X = X_L - X_C$ becomes zero. The impedance of the circuit is then purely resistive: $Z = R$.

   Since the impedance has no imaginary part, the voltage and current must rise and fall together — there is no lag or lead. The phase angle $\phi$ between voltage and current is given by $\tan \phi = \frac{X_L - X_C}{R}$. At resonance, $X_L - X_C = 0$, so $\tan \phi = 0$, which gives $\phi = 0$.

   Let’s walk through the reasoning step by step.

   1. Write the general expression for phase difference. In any series LCR circuit, the phase angle $\phi$ between the applied voltage and the current is:

   $\tan \phi = \frac{X_L - X_C}{R}$

   where $X_L = \omega L$, $X_C = \frac{1}{\omega C}$, and $R$ is the resistance.

   2. Apply the resonance condition. At resonance, the angular frequency $\omega_0$ satisfies:

   $\omega_0 L = \frac{1}{\omega_0 C}$

   So $X_L = X_C$, and therefore $X_L - X_C = 0$.

   3. Substitute into the phase formula. $\tan \phi = \frac{0}{R} = 0$ …

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9. COMEDK 2023Set 2023-M1 markMCQ
   Q.The phase difference between displacement and velocity of a particle in simple harmonic motion is (A) $\pi \mathrm{~rad}$ (B) $3 \pi / 2 \mathrm{~rad}$ (C) zero (D) $\pi / 2 \mathrm{~rad}$

   ›Reveal solutionSolution

   Velocity is the time-derivative of displacement in SHM, and differentiation shifts the phase by $\tfrac\pi2$; hence the phase difference is $\pi/2$.

   Let $x=A\sin\omega t$. Then:

   $v=\frac{dx}{dt}=A\omega\cos\omega t=A\omega\sin\!\left(\omega t+\frac{\pi}{2}\right).$ …

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10. COMEDK 2022Set 20221 markMCQ
    Q.During the phenomenon of resonance (A) the amplitude of oscillation becomes large (B) the frequency of oscillation becomes large (C) the time period of oscillation becomes large (D) All of the above

    ›Reveal solutionSolution

    The frequency of oscillation is set by the driver (it does not "become large"), and the time period likewise does not grow — it simply equals the natural period. So only (A) is correct.

    Concept: Resonance in a driven oscillator.

    Resonance occurs when the driving frequency equals (or is very near) the system's natural frequency. At that point the denominator of the steady-state amplitude,

    A = F₀/m ÷ √[(ω₀² − ω²)² + (bω/m)²],

    becomes minimal (only the damping term survives), so the amplitude of oscillation becomes very large. …

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11. COMEDK 2022Set 20221 markMCQ
    Q.A particle is performing simple harmonic motion. Equation of its motion is $x = 5\sin \left( {4t - {\pi \over 6}} \right),x$ being the displacement from mean position. Velocity (in ms$^{-1}$) of the particle at the instant when its displacement is 3, will be (A) ${{2\pi } \over 3}$ (B) ${{5\pi } \over 6}$ (C) 20 (D) 16

    ›Reveal solutionSolution

    At x = 3: v = 4 √(5² − 3²) = 4 √(25 − 9) = 4 √16 = 4 × 4 = 16 m/s.

    Concept: For SHM, v = ω√(A² − x²).

    From x = 5 sin(4t − π/6): amplitude A = 5, angular frequency ω = 4 rad/s.

    At x = 3: …

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12. COMEDK 2021Set 20211 markMCQ
    Q.Two spring of force constant $k_1$ and $k_2$ are configured as the figure given below The angular frequency of this configuration is (A) ${{{k_1} + {k_2}} \over m}$ (B) $\sqrt {{{{k_1} + {k_2}} \over m}}$ (C) $\sqrt {{{{k_1}} \over {{k_2}m}}}$ (D) ${{{k_2}m} \over {{k_1}}}$

    ›Reveal solutionSolution

    (Option A is dimensionally wrong for an angular frequency.)

    Concept: springs on OPPOSITE sides of a block, each fixed to a wall, act in PARALLEL (both are stretched/compressed by the same displacement x of the block, and both push/pull it back).

    From the figure: wall - k1 - block(m) - k2 - wall, all along one horizontal line.

    Displace the block by x to the right: spring 1 stretches and pulls back with k1x, spring 2 compresses and pushes back with k2x. Both restoring forces add: …

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