Q.A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of $50\ \text{N m}^{-1}$. The block is pulled to a distance $x = 10$ cm from its equilibrium position at $x = 0$ on a frictionless surface from rest at $t = 0$. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position.

Showing the 12 most recent of 44 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A particle is executing simple harmonic motion with an amplitude of 25 cm. If the kinetic energy of the particle at the mean position is 5 J, then the maximum force acting on the particle in its motion is (A) 20 N (B) 40 N (C) 50 N (D) 100 N

   ›Reveal solutionSolution

   This tests the SHM energy-force relation: at the mean position all energy is kinetic (giving total $E$), and the maximum restoring force at the extremes can be written directly in terms of $E$ and amplitude without ever needing $m$ or $\omega$ separately. Answer: (B) 40 N.

   Concept and Intuition

   In SHM, the restoring force is $F=-m\omega^2x$, maximum in magnitude at the extreme positions ($x=\pm A$): $F_{max}=m\omega^2A$. The total mechanical energy is constant throughout the motion and equals the kinetic energy at the mean position (where potential energy is zero): $E=\tfrac12mv_{max}^2=\tfrac12m\omega^2A^2$. Both expressions share the combination $m\omega^2$, so we can eliminate it between the two equations without knowing $m$ or $\omega$ individually — a common trick in SHM energy problems.

   Step-by-Step Solution

   1. Energy at mean position = total energy: $E=\tfrac12m\omega^2A^2=5\ J$, so

   $m\omega^2=\frac{2E}{A^2}=\frac{2(5)}{(0.25)^2}=\frac{10}{0.0625}=160\ N/m.$

   2. Maximum force at the extreme: $F_{max}=m\omega^2A=160\times0.25=40\ N.$ …

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2. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The mechanical energy of a damped harmonic oscillator at a time $t=0$ is 80 J. If its mechanical energy becomes 40 J after completing 15 oscillations, then the extra oscillations the oscillator has to complete so that its mechanical energy becomes 10 J is (A) 15 (B) 60 (C) 45 (D) 30

   ›Reveal solutionSolution

   A damped oscillator's energy decays exponentially with time (and hence with the number of completed oscillations, since each takes the same period) — use the given 15-oscillation decay ratio to project how many more oscillations are needed to reach a further energy level. Answer: (D) 30.

   Concept and Intuition

   For a lightly damped harmonic oscillator, $E(t)=E_0e^{-\gamma t}$, an exponential decay. Since each oscillation takes the same time period $T$ (weak damping barely changes the period), the energy after $n$ completed oscillations is $E_n=E_0\,r^n$ where $r=e^{-\gamma T}$ is a constant ratio — the energy shrinks by the same multiplicative factor every single oscillation, regardless of how much has already decayed. This is the hallmark of exponential decay: equal ratios over equal intervals.

   Step-by-Step Solution

   1. Find the per-oscillation decay factor from the given data: after 15 oscillations, $E_{15}=E_0r^{15}=40\ J$ with $E_0=80\ J$, so

   $r^{15}=\frac{40}{80}=\frac12.$

   2. Set up the target: we want the total number of oscillations $N=15+n$ (where $n$ is the extra oscillations asked for) such that $E_N=10\ J$:

   $E_0r^{N}=10 \Rightarrow r^{N}=\frac{10}{80}=\frac18.$

   3. Relate to the known factor: $r^{N}=r^{15+n}=r^{15}\cdot r^n=\frac12\cdot r^n$. Setting this equal to $\frac18$: $\frac12\cdot r^n=\frac18 \Rightarrow r^n=\frac14.$ …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.A particle is executing simple harmonic motion with amplitude $A$. The position at which kinetic energy and potential energy are equal is given by (A) $A/2$ (B) $2A$ (C) $A$ (D) $A/\sqrt{2}$

   ›Reveal solutionSolution

   In SHM, kinetic and potential energy are equal at a displacement of $A/\sqrt{2}$ from the mean position — exactly where potential energy is half the total energy.

   Concept and Intuition

   In SHM, total mechanical energy stays constant and continuously trades off between kinetic and potential forms. Setting KE = PE means each must be exactly half the total energy, which pins down a specific displacement independent of the SHM's frequency or mass — it only depends on the amplitude.

   Step-by-Step Solution

   1. Total energy in SHM: $E = \frac{1}{2}kA^2$ (constant, at either extreme where all energy is potential).
   2. At displacement $x$: $PE(x) = \frac{1}{2}kx^2$, and $KE(x) = E - PE(x) = \frac{1}{2}k(A^2-x^2)$.
   3. Setting $KE = PE$: $\frac{1}{2}k(A^2-x^2) = \frac{1}{2}kx^2$ …

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4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The springs are connected to the blocks as shown in figures A and B. When the blocks are slightly displaced and released, they oscillate with time period $T_A$ and $T_B$ respectively. Then, the value of $\dfrac{T_A}{T_B}$ is [FIGURE: In (A), a block of mass $3m$ is connected between two rigid walls by two springs each of spring constant $K$, one on either side. In (B), a block of mass $m$ is connected between two rigid walls by a spring of spring constant $K$ on one side and a spring of spring constant $2K$ on the other side.] (A) $\dfrac{3}{\sqrt2}$ (B) $\dfrac{\sqrt3}{\sqrt2}$ (C) $\dfrac{\sqrt2}{\sqrt3}$ (D) $\dfrac{\sqrt2}{3}$

   ›Reveal solutionSolution

   This tests the effective spring constant for two springs attached to a block from opposite fixed walls (springs in parallel add), then compares SHM periods for two such setups. The answer is (A), $3/\sqrt2$.

   Concept and Intuition

   When a block sits between two walls, connected by a spring on each side, a displacement of the block stretches one spring and compresses the other — both springs then push/pull the block back toward equilibrium in the same direction. This is the parallel-spring configuration, where the effective stiffness is the SUM of the individual spring constants: $k_{eff}=k_1+k_2$. The period of the resulting SHM is $T=2\pi\sqrt{m/k_{eff}}$.

   Step-by-Step Solution

   1. In setup (A): block of mass $3m$ has a spring of constant $K$ on each side (both attached to walls). Effective stiffness: $k_A=K+K=2K$.
   2. Period: $T_A=2\pi\sqrt{\dfrac{3m}{2K}}$.
   3. In setup (B): block of mass $m$ has spring $K$ on one side and spring $2K$ on the other. Effective stiffness: $k_B=K+2K=3K$.
   4. Period: $T_B=2\pi\sqrt{\dfrac{m}{3K}}$. …

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5. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.In the arrangement shown, $k_1= 1500\ Nm^{-1}$, $k_2= 500\ Nm^{-1}$, $m_1= 2\ kg$ and $m_2=1\ kg$. [FIGURE] (a fixed rigid support at the top, connected vertically downward in series to spring $k_1$, then mass $m_1$, then spring $k_2$, then mass $m_2$ hanging at the bottom) The potential energy stored in the system of springs in equilibrium is (Assume springs mass is negligible and $g= 10\ ms^{-2}$) (A) 0.5 J (B) 0.4 J (C) 1.2 J (D) 5.6 J

   ›Reveal solutionSolution

   This tests series-spring equilibrium: each spring stretches only under the weight it directly carries below it. Total elastic PE $= 0.4\ J$.

   Concept and Intuition

   When springs are connected in series with masses hanging in between (ceiling – $k_1$ – $m_1$ – $k_2$ – $m_2$), the tension in each spring is determined by everything hanging below that spring, not by the whole system's weight. This is unlike two springs simply in series holding one mass, where both springs carry the same tension — here different masses are inserted between the springs, so each spring carries a different load.

   Step-by-Step Solution

   1. The lowest spring $k_2$ connects $m_1$ (above) to $m_2$ (below). In equilibrium, the only mass hanging from $k_2$ is $m_2$, so the tension in $k_2$ is $F_2 = m_2 g = 1\times10 = 10\ N$.
   2. Extension of $k_2$: $x_2 = F_2/k_2 = 10/500 = 0.02\ m$.
   3. Elastic PE in $k_2$: $PE_2 = \tfrac12 k_2 x_2^2 = \tfrac12(500)(0.02)^2 = 0.1\ J$. …

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6. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.A particle of mass 40 gm is executing SHM. Variation of its potential energy ($E_P$) with square of its displacement is shown in the figure. Its time period of oscillation in seconds is: [FIGURE] (graph of $E_P$ (mJ) on the vertical axis against $x^2$ ($\mathrm{cm^2}$) on the horizontal axis; a straight line through the origin passing through the point $x^2=4,\ E_P=80$) (A) $\dfrac{\pi}{20}$ (B) $\dfrac{\pi}{50}$ (C) $\dfrac{\pi}{25}$ (D) $\dfrac{\pi}{100}$

   ›Reveal solutionSolution

   The slope of $E_P$ vs $x^2$ equals $\tfrac12 k$, giving $k=400\,\text{N/m}$; with $m=0.04\,\text{kg}$, $T=2\pi\sqrt{m/k}=\dfrac{\pi}{50}\,\text{s}$.

   For SHM the potential energy is $E_P=\tfrac12 k x^2$, so a plot of $E_P$ against $x^2$ is a straight line through the origin with slope $\tfrac12 k$.

   Read the slope from the point $x^2=4\,\text{cm}^2$, $E_P=80\,\text{mJ}$. Convert to SI: $x^2=4\times10^{-4}\,\text{m}^2$, $E_P=80\times10^{-3}\,\text{J}$.

   $\frac{1}{2}k=\frac{E_P}{x^2}=\frac{0.080}{4\times10^{-4}}=200\;\Rightarrow\;k=400\,\text{N/m}.$ …

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7. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The kinetic energy of a particle executing simple harmonic motion at a displacement of 3 cm from the mean position is 4 mJ. If the amplitude of the particle is 5 cm, then the maximum force acting on the particle is (A) 0.25 N (B) 0.50 N (C) 0.75 N (D) 1.25 N

   ›Reveal solutionSolution

   Using the SHM kinetic-energy formula to first find $m\omega^2$, then applying it to the amplitude gives the maximum restoring force: (A) 0.25 N.

   Concept and Intuition

   In SHM, the restoring force at displacement $x$ is $F=m\omega^2x$, which is maximum at the extreme position $x=A$ (the amplitude): $F_{max}=m\omega^2A$. We aren't given $m$ and $\omega$ separately, but the kinetic energy at a known displacement lets us solve for the combination $m\omega^2$ directly, which is exactly what we need for the force formula — we never need $m$ and $\omega$ individually.

   Step-by-Step Solution

   1. KE in SHM at displacement $x$: $KE=\tfrac12m\omega^2(A^2-x^2)$ (since total energy $\tfrac12m\omega^2A^2$ splits between KE and PE $=\tfrac12m\omega^2x^2$).
   2. Convert to consistent SI units: $x=3\ \mathrm{cm}=0.03$ m, $A=5\ \mathrm{cm}=0.05$ m, $KE=4\ \mathrm{mJ}=4\times10^{-3}$ J.
   3. Compute $A^2-x^2=0.0025-0.0009=0.0016\ \mathrm{m^2}$. …

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8. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.A body of mass 1 kg is suspended from a spring of force constant $600\ \mathrm{N\,m^{-1}}$. Another body of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of $3\ \mathrm{m\,s^{-1}}$ and embedded in it. The amplitude of motion is (A) 5 cm (B) 15 cm (C) 10 cm (D) 8 cm

   ›Reveal solutionSolution

   A perfectly inelastic collision shifts both the velocity (via momentum conservation) and the equilibrium point (via the new total weight); combining the resulting displacement-from-new-equilibrium and velocity through the SHM amplitude formula gives $A\approx5$ cm.

   Concept and Intuition

   Before collision, the 1 kg mass hangs in equilibrium at some extension $x_1=\dfrac{m_1g}{k}$. The 0.5 kg mass hits and embeds (perfectly inelastic collision), so momentum is conserved at that instant, giving the combined mass a velocity. But the combined mass's own equilibrium position (extension $x_2=\dfrac{(m_1+m_2)g}{k}$) is different (lower) from where the collision happened — so relative to its own new equilibrium, the combined body starts displaced by $\Delta x = x_2-x_1$, in addition to having the collision velocity. Both this displacement and this velocity feed into the amplitude of the resulting SHM.

   Step-by-Step Solution

   1. Momentum conservation at collision: $m_2 u = (m_1+m_2)v_0 \Rightarrow 0.5\times3=1.5\times v_0 \Rightarrow v_0=1\ \mathrm{m/s}$.
   2. Old equilibrium extension: $x_1=\dfrac{1\times10}{600}=\dfrac{1}{60}$ m. New equilibrium extension: $x_2=\dfrac{1.5\times10}{600}=\dfrac{1}{40}$ m.
   3. Displacement from new equilibrium at the moment of collision: $y_0=x_2-x_1=\dfrac{1}{40}-\dfrac{1}{60}=\dfrac{3-2}{120}=\dfrac{1}{120}\ \mathrm{m}$. …

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9. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.When the mass attached to a spring is increased from 4 kg to 9 kg, the time period of oscillation increases by $0.2\pi$ s. Then the spring constant of the spring is (A) $80\ \mathrm{N\,m^{-1}}$ (B) $200\ \mathrm{N\,m^{-1}}$ (C) $50\ \mathrm{N\,m^{-1}}$ (D) $100\ \mathrm{N\,m^{-1}}$

   ›Reveal solutionSolution

   Uses the spring-mass period formula twice to eliminate $k$ from a known period difference; $k=100\ \mathrm{N\,m^{-1}}$.

   Concept and Intuition

   The period of a mass-spring oscillator is $T=2\pi\sqrt{m/k}$ — it depends on $\sqrt{m}$, so equal jumps in mass don't give equal jumps in period, but the algebra stays linear in $\sqrt{k}$ once you plug in the actual masses (which happen to be perfect squares, 4 and 9), making $\sqrt{m}$ a clean integer.

   Step-by-Step Solution

   1. $T_1 = 2\pi\sqrt{\dfrac{4}{k}} = \dfrac{4\pi}{\sqrt{k}}$.
   2. $T_2 = 2\pi\sqrt{\dfrac{9}{k}} = \dfrac{6\pi}{\sqrt{k}}$. …

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10. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.A body of mass 4 kg attached to a spring of force constant $64\ \mathrm{N\,m^{-1}}$ executes simple harmonic motion on a frictionless horizontal surface. The time period of oscillation is (A) $\dfrac{\pi}{3}\ \mathrm{s}$ (B) $\dfrac{\pi}{2}\ \mathrm{s}$ (C) $\pi\ \mathrm{s}$ (D) $\dfrac{3\pi}{2}\ \mathrm{s}$

    ›Reveal solutionSolution

    This tests the direct application of the spring-mass SHM time period formula. Answer: (B).

    Concept and Intuition

    A mass on a spring on a frictionless surface undergoes SHM with restoring force $F=-kx$, giving angular frequency $\omega=\sqrt{k/m}$ and time period $T=2\pi/\omega=2\pi\sqrt{m/k}$ — independent of amplitude.

    Step-by-Step Solution

    1. Time period formula: $T = 2\pi\sqrt{\dfrac{m}{k}}$.
    2. Substitute $m=4\ \text{kg}$, $k=64\ \text{N/m}$:

    $T = 2\pi\sqrt{\frac{4}{64}} = 2\pi\sqrt{\frac{1}{16}} = 2\pi\times\frac14 = \frac{\pi}{2}\ \text{s}.$

    Common Mistakes …

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11. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
    Q.The ratio of kinetic energy and total energy of a particle in simple harmonic motion, at a point where the displacement is 30% of its amplitude is (A) 91 : 100 (B) 49 : 100 (C) 81 : 100 (D) 51 : 100

    ›Reveal solutionSolution

    The KE-to-total-energy ratio in SHM is $1-(x/A)^2$; at 30% amplitude displacement this evaluates to 91:100.

    Concept and Intuition

    In simple harmonic motion, total mechanical energy is conserved and constant, split between kinetic and potential energy depending on displacement $x$ from the mean position: $KE = \tfrac12 m\omega^2(A^2-x^2)$, $PE=\tfrac12m\omega^2x^2$, and $TE=\tfrac12m\omega^2A^2$ (using $k=m\omega^2$). Since $TE$ is position-independent, the fraction of energy that is kinetic depends only on how far (as a fraction of amplitude) the particle is displaced.

    Step-by-Step Solution

    1. Write $\dfrac{KE}{TE} = \dfrac{A^2-x^2}{A^2} = 1-\left(\dfrac{x}{A}\right)^2$.
    2. Given $x = 0.3A$ (30% of amplitude): $\left(\dfrac{x}{A}\right)^2 = 0.09$. …

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12. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
    Q.The length of the scale of a spring balance that can weigh zero to 100 kg is 25 cm. If a body suspended from this balance oscillates with a time period of $0.2\pi$ s, then the mass of the suspended body is (Acceleration due to gravity = $10\ \mathrm{m\,s^{-2}}$) (A) 30 kg (B) 50 kg (C) 40 kg (D) 60 kg

    ›Reveal solutionSolution

    The scale's markings fix the spring constant ($k=4000$ N/m); using the SHM period formula with the given $T=0.2\pi$ s gives a suspended mass of 40 kg.

    Concept and Intuition

    A spring balance is just a calibrated spring: its full deflection (25 cm) corresponds to its maximum rated weight (100 kg). This directly gives the spring constant $k = F/x$. Once $k$ is known, any mass hung on it and set oscillating obeys the standard mass-spring SHM period formula $T = 2\pi\sqrt{m/k}$, from which the oscillating mass can be found.

    Step-by-Step Solution

    1. Full-scale force: $F = 100 \times 10 = 1000$ N, corresponding to extension $x = 0.25$ m.
    2. Spring constant: $k = \dfrac{F}{x} = \dfrac{1000}{0.25} = 4000\ \mathrm{N/m}$.
    3. SHM period of a mass $m$ on this spring: $T = 2\pi\sqrt{\dfrac{m}{k}}$. …

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