Q.Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?

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 $90^\circ$.

   Common Mistakes

   • Confusing "leads" and "lags" — since differentiating a sine gives a cosine which peaks earlier in the cycle, velocity leads (not lags) displacement.
   • Mixing this up with the acceleration-vs-displacement relationship, which is $180^\circ$ out of phase (acceleration is anti-parallel to displacement, $a=-\omega^2x$), not $90^\circ$.

   ✓Final answer

   The correct option is (B) — $90°$.

   ANSWER: B

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 $x(t) = A\cos(\omega t + \pi) = -A\cos(\omega t) = -3\cos(\pi t)$.
   7. Note this result does not depend on the stated sense (clockwise) of revolution: swapping the sense is the same as replacing $t$ by $-t$ in the reference circle's angle, and since $\cos$ is even, $-3\cos(-\pi t) = -3\cos(\pi t)$ — identical. The sense would only matter if the particle started at a point where the x-projection was passing through zero (not at an extremum).

   Common Mistakes

   • Writing the answer as a sine function because "SHM starts with sine" — that template only applies when the particle starts at the mean position ($x=0$), not at an extreme point.
   • Forgetting the sign: since the particle starts on the negative x-axis, the leading coefficient must be negative, ruling out $x=3\cos(\pi t)$ (which is not even offered) as well as $x = \pm 3\sin(\pi t)$.
   • Believing the clockwise/anticlockwise direction changes the x-projection when starting from an extreme point — it doesn't, only the y-projection (or the direction of initial velocity) would differ.

   ✓Final answer

   The correct option is (C) — $x = -3\cos(\pi t)$.

   ANSWER: C

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).
   4. Kinetic energy: since total energy $E=KE+PE$ is constant and $PE=0$ here, $KE=E$ — its maximum possible value.
   5. Hence at mean position, KE is maximum.

   Common Mistakes

   • Confusing mean position (KE max) with extreme position (PE max, acceleration max, KE zero).

   ✓Final answer

   The correct option is (A) — $KE$ maximum.

   ANSWER: A

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.
   6. The particle reverses direction only when $v_x=0$, i.e. $\omega t=\pi/2\Rightarrow t=1$ s — later than $0.5$ s — so no reversal occurs in this sub-interval, and the distance equals $|x(0.5)-x(0)|=9.91$ cm.
   7. Total distance $\approx56.05+9.91=65.96\approx65.9$ cm.

   Common Mistakes

   • Assuming $4.5$ s $= 1.125\,T$ and multiplying $4A\times1.125$ blindly — this overcounts because distance per unit time is NOT uniform across a period.
   • Forgetting to check whether the particle reverses direction inside the leftover fractional interval.

   ✓Final answer

   The correct option is (C) — 65.9 cm.

   ANSWER: C

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. $x = A\cos(30^\circ) = A\cdot\dfrac{\sqrt3}{2} = \dfrac{A\sqrt3}{2}$.
   5. Distance travelled from the extreme position: $\Delta x = A - x = A - \dfrac{A\sqrt3}{2} = \dfrac{A(2-\sqrt3)}{2}$.

   Common Mistakes

   • Using $x(t) = A\sin(\omega t)$, which is correct for starting from the mean position, not from rest at the extreme.
   • Forgetting to subtract from $A$ — the question asks for distance travelled, not the instantaneous position.

   ✓Final answer

   The correct option is (B) — $\dfrac{A(2-\sqrt{3})}{2}$.

   ANSWER: B

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: $v = \dfrac{dx}{dt} = 6(2\pi)\cos\left(2\pi t+\frac{\pi}{4}\right) = 12\pi\cos(\cdots)$.
   3. Maximum value of $\cos(\cdots)$ is 1, so $v_{max} = 12\pi\ ms^{-1}$.

   Common Mistakes

   • Misreading the amplitude as something derived from the phase constant $\pi/4$ (irrelevant to the amplitude/max-speed values).
   • Forgetting to multiply amplitude by $\omega$ (just stating $A$ as the max speed).

   ✓Final answer

   The correct option is (C) — $6\,m,\ 12\pi\ ms^{-1}$.

   ANSWER: C

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).
   4. Common denominator: $\dfrac{2\pi}{3}=\dfrac{4\pi}{6}$, $\dfrac{\pi}{2}=\dfrac{3\pi}{6}$, so the bracket is $\dfrac{4\pi}{6}-\dfrac{3\pi}{6}=\dfrac{\pi}{6}$.
   5. So $\Delta\phi(t)=\dfrac{\pi}{6}t$. At $t=1$ s: $\Delta\phi=\dfrac{\pi}{6}$.

   Common Mistakes

   • Forgetting that both expressions include the same $+\phi$, and mistakenly trying to compute a phase difference that depends on $\phi$ (it doesn't — it cancels).
   • Arithmetic slip converting $\pi/2$ and $2\pi/3$ to a common denominator.

   ✓Final answer

   The correct option is (D) — $\dfrac{\pi}{6}$.

   ANSWER: D

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.
   5. Maximum acceleration occurs at the extreme position: $a_{max}=\omega^2\cdot(\text{amplitude})=\dfrac{A}{B}\times\sqrt{\dfrac{E}{A}}=\dfrac{\sqrt{A}\sqrt{E}}{B}=\dfrac{\sqrt{AE}}{B}$. This matches option (D) — correct statement.
   6. Since (B), (C), (D) all check out against the derived relations but (A) has an extra unjustified factor of $\sqrt2$, (A) is the INCORRECT statement asked for.

   Common Mistakes

   • Confusing the amplitude formula with the maximum-velocity formula's structure (mistakenly inserting a factor of 2 as if amplitude used total energy differently than it does).
   • Not deriving all four options independently and instead assuming the odd-one-out by pattern rather than checking the algebra.

   ✓Final answer

   The correct option is (A) — Amplitude is $\sqrt{\dfrac{2E}{A}}$ (this is the INCORRECT statement; the true amplitude is $\sqrt{E/A}$).

   ANSWER: A

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: $\left(\dfrac{x}{A}\right)^2+\left(\dfrac{y}{B}\right)^2 = \sin^2\omega t + \cos^2\omega t = 1$.
   4. This is $\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1$ — the standard equation of an ellipse with semi-axes $A$ and $B$.
   5. Since $A \neq B$ is given, the ellipse is not a circle; the resultant motion is genuinely elliptical.

   Common Mistakes

   • Concluding "circular path" just because the frequencies are equal — equal frequency with $90^\circ$ phase difference gives a circle only if the amplitudes also match.
   • Trying to eliminate $t$ without first normalizing $x$ and $y$ by their respective amplitudes.

   ✓Final answer

   The correct option is (C) — Elliptical path.

   ANSWER: C

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: $\omega_{eff}^2=\dfrac{k_{eff}}{m}=4\pi^2\left(\dfrac1{T_1^2}+\dfrac1{T_2^2}\right)$.
    5. New period: $T_{eff}=\dfrac{2\pi}{\omega_{eff}}=\dfrac{2\pi}{2\pi\sqrt{1/T_1^2+1/T_2^2}}=\dfrac{1}{\sqrt{1/T_1^2+1/T_2^2}}$.
    6. Simplify: $\dfrac{1}{\sqrt{(T_1^2+T_2^2)/(T_1^2T_2^2)}}=\sqrt{\dfrac{T_1^2T_2^2}{T_1^2+T_2^2}}$.

    Common Mistakes

    • Adding the periods directly ($T_1+T_2$) — periods don't add when the underlying force constants (which are $\propto1/T^2$) add.
    • Adding angular frequencies instead of their squares (force constants) — this is a subtlety of "parallel spring" combination.

    ✓Final answer

    The correct option is (D) — $\sqrt{\dfrac{T_1^2T_2^2}{T_1^2+T_2^2}}\ sec$.

    ANSWER: D

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.
    5. Checking each option: (A) KE max at $x=0$ — TRUE. (B) TE zero at $x=0$ — false, TE is constant and non-zero. (C) KE max when $x$ is maximum — false, KE is zero there. (D) PE max at $x=0$ — false, PE is zero there.

    Common Mistakes

    • Confusing where KE and PE individually peak — students often swap them, thinking PE builds up at the centre.
    • Believing TE varies with $x$; in ideal SHM it is strictly constant.

    ✓Final answer

    The correct option is (A) — KE is maximum when $x = 0$.

    ANSWER: A

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 $t = \dfrac{\pi/3}{2\pi/T} = \dfrac{T}{6}$.
    5. Average velocity magnitude $= \dfrac{\text{distance travelled}}{\text{time taken}} = \dfrac{a/2}{T/6} = \dfrac{3a}{T}$.

    Common Mistakes

    • Using $x=a\sin(\omega t)$ (measured from the mean position) and then mis-locating "distance $a/2$ from extreme" — it's cleaner to measure phase from the extreme directly.
    • Confusing average velocity (distance/time) with average speed over a full cycle or with instantaneous velocity at a point.

    ✓Final answer

    The correct option is (D) — $\dfrac{3a}{T}$.

    ANSWER: D