Q.The displacement–time curve of a particle executing S.H.M. is a smooth sinusoid of period $4\ \text{s}$. The displacement is positive and near its maximum at $t=0\ \text{s}$, falls to zero at about $t=1\ \text{s}$, reaches its most negative value at about $t=2\ \text{s}$, returns to zero at about $t=3\ \text{s}$, and rises back to its positive maximum at $t=4\ \text{s}$; this 4-second cycle then repeats up to $t=7\ \text{s}$. Choose the correct statement(s). (Note: more than one of the given options may be correct.)

Showing the 12 most recent of 96 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.In SHM, velocity leads displacement by (A) 0 (B) $90°$ (C) $180°$ (D) $45°$

   ›Reveal solutionSolution

   Differentiating the SHM displacement gives velocity as a cosine of the same phase, which is exactly a sine shifted forward by $90^\circ$ — velocity leads displacement by $90^\circ$.

   Concept and Intuition

   In SHM, displacement, velocity, and acceleration are all sinusoidal with the same angular frequency $\omega$, but shifted in phase relative to each other because velocity is the time-derivative of displacement (and acceleration the derivative of velocity). Each derivative operation on a sinusoid produces a $90^\circ$ phase shift.

   Step-by-Step Solution

   1. Let displacement be $x = A\sin(\omega t)$.
   2. Velocity: $v = \dfrac{dx}{dt} = A\omega\cos(\omega t)$.
   3. Rewrite cosine as a phase-shifted sine: $\cos(\omega t) = \sin(\omega t + 90^\circ)$.
   4. So $v = A\omega\sin(\omega t+90^\circ)$ — this is displacement's waveform shifted ahead by $90^\circ$, i.e. velocity leads displacement by $90^\circ$. …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The figure corresponds to circular motion where the radius of the circle, the period of revolution, the initial position and sense of revolution (ie clockwise or anti-clockwise) are indicated. The corresponding simple harmonic motion of the x-projection of the radius vector of the revolving particle p is (Figure: a circle of radius 3 m centred at the origin of the x-y plane. The particle p starts at t = 0 at the leftmost point of the circle, i.e. on the negative x-axis. The period of revolution is T = 2 s. The sense of revolution is clockwise, as indicated by an arrow at the bottom of the circle.) (A) $x = 3\sin \pi t$ (B) $x = -3\sin \pi t$ (C) $x = -3\cos \pi t$ (D) $x = 2\sin\left(\dfrac{2\pi}{3}\right)t$

   ›Reveal solutionSolution

   The x-projection of uniform circular motion is SHM with the same $\omega$ as the rotation; starting at an extreme point of the projection fixes it as a (negative) cosine, giving $x=-3\cos(\pi t)$.

   Concept and Intuition

   When a particle moves uniformly on a circle of radius $A$, the projection of its position vector on any diameter executes SHM with amplitude $A$ and the same angular frequency $\omega$ as the circular motion. The phase of that SHM is fixed entirely by where the particle starts and how fast its projection is changing there — nothing more is needed once you know the starting point is a turning point (extremum) or a zero-crossing of the projection.

   Step-by-Step Solution

   1. Angular frequency: $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{2\,\text{s}} = \pi\ \text{rad/s}$.
   2. Amplitude of the x-projection equals the radius: $A = 3\ \text{m}$.
   3. At $t=0$ the particle $p$ is at the leftmost point of the circle, so its position is $(-3,0)$, i.e. $x(0) = -3\ \text{m} = -A$.
   4. At the leftmost/rightmost points of a circle, the velocity of the particle is purely tangential, i.e. purely along $y$; the instantaneous rate of change of $x$ is zero there. So $x=0$ point is a turning point of the SHM (like the extreme ends of a spring's oscillation), not a point where $x$ passes through zero.
   5. A general SHM projection is $x(t) = A\cos(\omega t + \phi)$. Turning points of $\cos$ occur at $t=0$ automatically when $\phi = 0$ or $\pi$; we need $x(0) = -A$, so $\cos\phi = -1 \Rightarrow \phi = \pi$.
   6. Hence $x(t) = A\cos(\omega t + \pi) = -A\cos(\omega t) = -3\cos(\pi t)$. …

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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.In simple harmonic motion, at mean position (A) $KE$ maximum (B) $PE$ maximum (C) $KE$ zero (D) Acceleration maximum

   ›Reveal solutionSolution

   Tests the basic energy/acceleration profile of SHM at the mean position. Answer: KE is maximum there.

   Concept and Intuition

   In SHM, total energy $E=\frac12kA^2$ is conserved and continuously exchanged between kinetic and potential forms as the particle oscillates. At the mean position (equilibrium, $x=0$), the restoring force and hence acceleration are zero, so the particle has been accelerating toward this point the entire half-swing — meaning its speed, and hence KE, peaks exactly here.

   Step-by-Step Solution

   1. Displacement from mean position: $x=0$ at mean position.
   2. Potential energy: $PE=\frac12kx^2=0$ at $x=0$ — minimum (zero), not maximum.
   3. Acceleration: $a=-\omega^2x=0$ at $x=0$ — zero, not maximum (acceleration is maximum at extreme positions). …

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4. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.A particle starting from the origin executes simple harmonic motion along x-axis. Its velocity at any instant $t$ is given by $v_x = 22\cos\left(\dfrac{\pi t}{2}\right)\ cm\,s^{-1}$. The total distance covered by the particle in time $t = 4.5\ sec$ (A) 74.5 cm (B) 51.4 cm (C) 65.9 cm (D) 49.8 cm

   ›Reveal solutionSolution

   Tests computing total distance in SHM over a non-integer number of periods using the velocity equation. Answer: $\approx65.9$ cm.

   Concept and Intuition

   Since $v_x=22\cos(\pi t/2)$ is maximum at $t=0$, the particle starts at the mean position ($x=0$) moving with maximum speed — this fixes $x(t)=A\sin(\omega t)$ with $A\omega = 22$. Total distance traveled per full period is $4A$ (since the particle sweeps mean→extreme→mean→extreme→mean, covering $A$ four times). For a leftover fractional time, we track where the particle actually is and add only the distance covered in that leftover interval, being careful whether it reverses direction within it.

   Step-by-Step Solution

   1. Compare $v_x=22\cos(\pi t/2)$ with $v_x=A\omega\cos(\omega t)$: $\omega=\pi/2\ rad\,s^{-1}$ and $A\omega=22\Rightarrow A=\dfrac{22}{\pi/2}=\dfrac{44}{\pi}\approx14.01$ cm.
   2. Period $T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{\pi/2}=4$ s.
   3. $t=4.5$ s $= 1\times T + 0.5$ s. In one full period the particle covers $4A\approx4(14.01)=56.05$ cm.
   4. Since $T=4$ s, the state (position and direction) at $t=4$ s is identical to $t=0$ s (mean position, moving in $+x$). So the extra $0.5$ s behaves exactly like the interval $t=0\to0.5$ s.
   5. Displacement: $x(t)=A\sin(\omega t)$ (zero at $t=0$, matching the cosine velocity). At $t=0.5$: $x=A\sin(\pi/4)=14.01\times0.7071\approx9.91$ cm. …

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5. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.A particle is executing simple harmonic motion with time period $T$ and amplitude $A$. The distance travelled by the particle in $\dfrac{T}{12}$ time starting from rest is (A) $\dfrac{A(\sqrt{3}-\sqrt{2})}{2}$ (B) $\dfrac{A(2-\sqrt{3})}{2}$ (C) $\dfrac{2A}{(2-\sqrt{3})}$ (D) $\dfrac{2A}{(\sqrt{3}-\sqrt{2})}$

   ›Reveal solutionSolution

   This tests SHM position as a function of time when starting from rest (i.e., from an extreme position). The answer is $\dfrac{A(2-\sqrt{3})}{2}$.

   Concept and Intuition

   "Starting from rest" in SHM means the particle begins at an extreme (turning) point, where velocity is momentarily zero and displacement is maximum ($x = A$). The position as a function of time measured from that instant is $x(t) = A\cos(\omega t)$, since $\cos(0) = 1$ gives $x=A$ at $t=0$ and velocity $\propto -\sin(\omega t) = 0$ at $t=0$. The distance travelled toward the mean position is $A - x(t)$.

   Step-by-Step Solution

   1. Angular frequency: $\omega = \dfrac{2\pi}{T}$.
   2. Position from extreme: $x(t) = A\cos(\omega t)$.
   3. At $t = \dfrac{T}{12}$: $\omega t = \dfrac{2\pi}{T}\times\dfrac{T}{12} = \dfrac{\pi}{6}$ (i.e. $30^\circ$).
   4. $x = A\cos(30^\circ) = A\cdot\dfrac{\sqrt3}{2} = \dfrac{A\sqrt3}{2}$. …

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6. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The displacement of particle executing simple harmonic motion is given by $x = 6\sin\left(2\pi t + \frac{\pi}{4}\right)\ m$. The amplitude and maximum speed are given by (A) $4\,m,\ 2\pi\ ms^{-1}$ (B) $6\,m,\ 4\pi\ ms^{-1}$ (C) $6\,m,\ 12\pi\ ms^{-1}$ (D) $2\,m,\ 12\pi\ ms^{-1}$

   ›Reveal solutionSolution

   Reading amplitude and angular frequency directly off the SHM equation and using $v_{max}=A\omega$ gives amplitude 6 m and max speed $12\pi\ ms^{-1}$.

   Concept and Intuition

   Any SHM displacement equation of the form $x = A\sin(\omega t + \phi)$ directly reveals the amplitude $A$ and angular frequency $\omega$ by inspection. The maximum speed in SHM always occurs while passing through the mean position, and its magnitude is $A\omega$ (from differentiating $x$ with respect to time and taking the maximum value of the cosine term).

   Step-by-Step Solution

   1. Given $x = 6\sin\left(2\pi t + \frac{\pi}{4}\right)$, matching to $x = A\sin(\omega t+\phi)$: $A = 6\ m$, $\omega = 2\pi\ rad/s$.
   2. Velocity: $v = \dfrac{dx}{dt} = 6(2\pi)\cos\left(2\pi t+\frac{\pi}{4}\right) = 12\pi\cos(\cdots)$. …

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7. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.Two Simple Harmonic Motions are given by $y_1 = A\sin\left(\dfrac{\pi}{2}t + \phi\right)$ and $y_2 = B\sin\left(\dfrac{2\pi}{3}t + \phi\right)$. The phase difference between these two after one sec is (A) $\pi$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   The two SHMs share the same initial phase $\phi$ but different angular frequencies; subtracting their phase expressions and evaluating at $t=1$ s gives a phase difference of $\pi/6$ — option (D).

   Concept and Intuition

   Both oscillations start with the same phase constant $\phi$, but because their angular frequencies differ ($\pi/2$ vs $2\pi/3$), the two phases drift apart as time passes. The phase difference at any instant is simply the difference of the two full phase expressions (the $\phi$ cancels out since both share it).

   Step-by-Step Solution

   1. Phase of $y_1$: $\phi_1(t)=\dfrac{\pi}{2}t+\phi$.
   2. Phase of $y_2$: $\phi_2(t)=\dfrac{2\pi}{3}t+\phi$.
   3. Phase difference: $\Delta\phi(t)=\phi_2(t)-\phi_1(t)=\left(\dfrac{2\pi}{3}-\dfrac{\pi}{2}\right)t$ (the $\phi$ terms cancel). …

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8. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The energy of a particle executing Simple Harmonic Motion (SHM) is given by $E = Ax^2 + BV^2$. Here 'x' is the displacement of the particle from mean position and 'V' is its velocity at 'x' and A and B are constants. The INCORRECT statement is (A) Amplitude is $\sqrt{\dfrac{2E}{A}}$ (B) Maximum velocity is $\sqrt{\dfrac{E}{B}}$ (C) Time period is $2\pi\sqrt{\dfrac{B}{A}}$ (D) Maximum acceleration is $\dfrac{\sqrt{EA}}{B}$

   ›Reveal solutionSolution

   This tests deriving amplitude, maximum velocity, time period, and maximum acceleration from a given SHM energy expression $E=Ax^2+BV^2$, and spotting which standard-looking statement is wrong. The answer is (A) — the amplitude formula given is off by a factor of $\sqrt2$.

   Concept and Intuition

   In SHM, total mechanical energy is conserved and equals $E=\tfrac12 m\omega^2x^2+\tfrac12 mV^2$ at any instant (PE + KE). By matching this to the given form $E=Ax^2+BV^2$, we identify $A=\tfrac12m\omega^2$ and $B=\tfrac12m$, and can then derive amplitude, max velocity, time period, and max acceleration purely from $A$, $B$, and $E$ — checking each option against the true SHM relations.

   Step-by-Step Solution

   1. Standard SHM energy: $E=\tfrac12 m\omega^2x^2+\tfrac12 mV^2$ (constant total energy). Matching coefficients with $E=Ax^2+BV^2$: $A=\tfrac12m\omega^2$, $B=\tfrac12m$. So $m=2B$ and $\omega^2=\dfrac{2A}{m}=\dfrac{2A}{2B}=\dfrac{A}{B}$.
   2. Maximum velocity occurs at $x=0$ (mean position), where all energy is kinetic: $E=BV_{max}^2\Rightarrow V_{max}=\sqrt{E/B}$. This matches option (B) — correct statement.
   3. Amplitude $a$ occurs where $V=0$ (extreme position), where all energy is potential: $E=Aa^2\Rightarrow a=\sqrt{E/A}$. Option (A) states the amplitude is $\sqrt{2E/A}$ — this has an extra factor of $\sqrt2$ that shouldn't be there, so option (A) is WRONG.
   4. Time period: $T=\dfrac{2\pi}{\omega}=2\pi\sqrt{\dfrac{1}{\omega^2}}=2\pi\sqrt{\dfrac{B}{A}}$ (since $\omega^2=A/B$). This matches option (C) — correct statement. …

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9. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.A particle executes two simple harmonic motions along mutually perpendicular axes, given by $x=A\sin(\omega_1 t)$, $y=B\cos(\omega_2 t)$ where $A\neq B$ and $\omega_1=\omega_2$. Which of the following best describes the resultant motion of the particle? (A) Straight line (B) Circular path (C) Elliptical path (D) Spiral path

   ›Reveal solutionSolution

   Two perpendicular SHMs of equal frequency with a $90^\circ$ phase difference trace an ellipse in general, collapsing to a circle only in the special case of equal amplitudes. Here $A\neq B$, so the path is elliptical.

   Concept and Intuition

   Combining perpendicular SHMs is a classic way to generate Lissajous figures. When the two frequencies are equal and the phase difference is exactly $90^\circ$ (as $\sin$ and $\cos$ are), the resulting path is always a closed ellipse whose semi-axes equal the two component amplitudes. Only when those amplitudes happen to be equal does the ellipse become a perfect circle. Since we're told $A \ne B$, the shape must remain a genuine ellipse.

   Step-by-Step Solution

   1. Given $x = A\sin(\omega_1 t)$ and $y = B\cos(\omega_2 t)$ with $\omega_1=\omega_2=\omega$.
   2. Rewrite: $\dfrac{x}{A} = \sin\omega t$ and $\dfrac{y}{B} = \cos\omega t$.
   3. Square and add: $\left(\dfrac{x}{A}\right)^2+\left(\dfrac{y}{B}\right)^2 = \sin^2\omega t + \cos^2\omega t = 1$. …

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10. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
    Q.A body executes SHM under the influence of one force and has the time period of $T_1$ seconds. The same body executes SHM with a time period of $T_2$ seconds when under the influence of another force separately. When both forces act simultaneously and in the same direction, then the time period of the same body is (A) $(T_1 + T_2)\ sec$ (B) $\sqrt{T_1^2 + T_2^2}\ sec$ (C) $\sqrt{\dfrac{T_1+T_2}{T_1 T_2}}\ sec$ (D) $\sqrt{\dfrac{T_1^2 T_2^2}{T_1^2+T_2^2}}\ sec$

    ›Reveal solutionSolution

    When two SHM-producing forces act together, their effective force constants (each $\propto1/T^2$) add, giving a combined period $T=\sqrt{\dfrac{T_1^2T_2^2}{T_1^2+T_2^2}}$.

    Concept and Intuition

    A restoring force that produces SHM behaves like a spring with some effective force constant $k=m\omega^2$. If two such restoring forces act on the same body simultaneously and in the same sense (both pulling toward the same equilibrium), their force constants simply add — exactly like two springs in parallel — giving a stiffer combined restoring force and hence a different (shorter) period.

    Step-by-Step Solution

    1. For a body of mass $m$ under one force alone with period $T_1$: $\omega_1=\dfrac{2\pi}{T_1}$, so effective force constant $k_1=m\omega_1^2=\dfrac{4\pi^2m}{T_1^2}$.
    2. Similarly for the second force alone: $k_2=\dfrac{4\pi^2m}{T_2^2}$.
    3. When both forces act simultaneously in the same direction (both restoring toward the same point), they combine like parallel springs: $k_{eff}=k_1+k_2=4\pi^2m\left(\dfrac1{T_1^2}+\dfrac1{T_2^2}\right)$.
    4. New angular frequency: $\omega_{eff}^2=\dfrac{k_{eff}}{m}=4\pi^2\left(\dfrac1{T_1^2}+\dfrac1{T_2^2}\right)$. …

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11. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
    Q.For a body in simple harmonic motion, Potential energy (PE), Kinetic energy (KE), Total energy (TE) are measured as a function of displacement x. Which of the following statements is true? (A) KE is maximum when $x = 0$ (B) TE is zero when $x = 0$ (C) KE is maximum when x is maximum (D) PE is maximum when $x = 0$

    ›Reveal solutionSolution

    Tests how KE, PE and TE vary with displacement in SHM — KE peaks at the mean position, PE peaks at the extremes, and TE is constant.

    Concept and Intuition

    A particle in SHM trades energy continuously between kinetic and potential forms while the total mechanical energy stays fixed (no non-conservative forces). Speed is greatest as the particle passes through the mean position (it has "picked up" all the energy that was stored as potential energy at the extremes) and speed is zero exactly at the extremes, where all the energy is potential.

    Step-by-Step Solution

    1. For SHM, $x = a\sin(\omega t)$, so $v = a\omega\cos(\omega t)$, giving $v^2 = \omega^2(a^2-x^2)$.
    2. $KE = \tfrac12 mv^2 = \tfrac12 m\omega^2(a^2-x^2)$. This is maximum when $x=0$ (since $a^2-x^2$ is largest there) and zero when $x=\pm a$.
    3. $PE = \tfrac12 m\omega^2 x^2$ is the complement: zero at $x=0$, maximum at $x=\pm a$.
    4. $TE = KE+PE = \tfrac12 m\omega^2 a^2$, independent of $x$ — never zero (for $a\neq0$) and never varies with position. …

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12. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
    Q.A particle executes simple harmonic motion with time period 'T' and amplitude 'a'. The magnitude of average velocity of the particle over the time interval during which it travels a distance $\dfrac{a}{2}$ from extreme position is (A) $\dfrac{a}{T}$ (B) $\dfrac{2a}{T}$ (C) $\dfrac{a}{2T}$ (D) $\dfrac{3a}{T}$

    ›Reveal solutionSolution

    Tests average velocity (distance/time, not instantaneous velocity) for a specific segment of SHM measured from the extreme position.

    Concept and Intuition

    Average velocity over an interval is just total distance covered divided by the time taken — it doesn't need calculus once we know the time. Measuring the phase from the extreme position (where velocity is zero) makes the position-time relation a cosine, which is the convenient choice here.

    Step-by-Step Solution

    1. Take the extreme position as the reference: $x(t) = a\cos(\omega t)$, with $\omega = \dfrac{2\pi}{T}$.
    2. The particle has travelled a distance $a/2$ from the extreme, so its new position (measured from the centre) is $x = a - \dfrac{a}{2} = \dfrac{a}{2}$.
    3. Set $a\cos(\omega t) = \dfrac{a}{2} \Rightarrow \cos(\omega t) = \dfrac12 \Rightarrow \omega t = \dfrac{\pi}{3}$.
    4. So $t = \dfrac{\pi/3}{2\pi/T} = \dfrac{T}{6}$. …

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