📌 Snapshot
- Learn simple harmonic motion from scratch: formulas, energy, springs and pendulums with solved CUET problems.
- Why SHM is the projection of circular motion, and how that one idea gives you x, v and a without calculus.
- The spring and the pendulum compared side by side, with the traps NTA sets on springs, lifts and the Moon.
- Six worked examples, an exam traps table and a formula sheet you can revise from in a minute.
📖 Detailed Notes
Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XI, Chapter 13.
Why does a swing keep the same rhythm?
Push a child on a swing gently and then harder. The swing goes higher, but the time for one to-and-fro trip barely changes. Pluck a guitar string softly or loudly and the note has the same pitch. That strange fact, the rhythm does not depend on how big the motion is, is the heart of this chapter. It happens only for one special kind of motion, called simple harmonic motion (SHM), and CUET loves it because a single idea gives you a dozen formulas.
Periodic, oscillatory and simple harmonic: three different things
Periodic motion repeats itself after equal intervals of time. Oscillatory motion is a periodic motion that goes to and fro about a mean (equilibrium) position. So every oscillation is periodic, but not every periodic motion oscillates. The Earth going round the Sun is periodic, but it does not move to and fro about a point.
Simple harmonic motion is the oscillation whose displacement is a pure cosine (or sine) of time:
$$x(t) = A\cos(\omega t + \phi)$$
The picture to hold in your head is a block on a spring or a bob on a string. Whatever the system, one test decides whether it is SHM: the acceleration must be proportional to the displacement and always point towards the mean position, $a = -\omega^{2}x$. A bouncing ball is periodic, yet its acceleration is constant (g), so it is not SHM.
| Term | Meaning | Unit |
|---|---|---|
| Displacement x | Distance from the mean position, positive on one side and negative on the other. In this chapter it can also be an angle, a voltage or a pressure | m |
| Amplitude A | Largest displacement. Always taken positive | m |
| Period T | Smallest time after which the motion repeats | s |
| Frequency $\nu$ | Oscillations per second, $\nu = 1/T$ | Hz (s⁻¹) |
| Angular frequency $\omega$ | $\omega = 2\pi/T = 2\pi\nu$. It tells how fast the reference point goes round, not how many oscillations per second | rad s⁻¹ |
| Phase ($\omega t + \phi$) | The angle that fixes the state of the motion at time t | rad |
| Phase constant $\phi$ | The phase at t = 0. It decides where the particle starts | rad |
Changing A stretches the graph up and down. Changing $\omega$ squeezes or stretches it sideways. Changing $\phi$ slides it left or right without changing its shape.
The big idea: SHM is a shadow of circular motion
Why is a cosine the right function? Imagine a ball going round a circle of radius A at constant angular speed $\omega$. Shine a light from the side so its shadow falls on a diameter. The shadow moves to and fro between +A and −A. If the ball starts at angle $\phi$, then at time t it is at angle $\omega t + \phi$, and the shadow is at
$$x = A\cos(\omega t + \phi)$$
That is exactly the SHM equation, so SHM is the projection of uniform circular motion on a diameter. The circle is called the reference circle. Now everything else follows by taking the same shadow of the ball's velocity and acceleration:
The ball's speed is $\omega A$, tangent to the circle. Its shadow gives $v = -A\omega \sin(\omega t + \phi)$. The ball's acceleration is $\omega^{2}A$, pointing at the centre. Its shadow gives $a = -A\omega^{2} \cos(\omega t + \phi) = -\omega^{2} x$.
Squaring and adding the x and v expressions removes time and gives the formula you will use most:
$$v = \pm \omega\sqrt{A^{2} - x^{2}}$$
Read it in words. At x = 0 the speed is largest, $v_{\max} = A\omega$. At x = ±A the speed is zero, because the particle stops to turn around. The acceleration is the opposite: zero at x = 0 and largest, $a_{\max} = \omega^{2}A$, at the ends.
| Position | x | Speed v | Acceleration a | Kinetic energy | Potential energy |
|---|---|---|---|---|---|
| Mean | 0 | Maximum, $A\omega$ | 0 | Maximum, all of E | 0 |
| Extreme | ±A | 0 | Maximum, $\omega^{2}A$ | 0 | Maximum, all of E |
| Quarter amplitude | $\pm A/2$ | $(\sqrt{3}/2) A\omega$ | $\omega^{2}A/2$ | $3E/4$ | $E/4$ |
| Where K = U | $\pm A/\sqrt{2}$ | $A\omega/\sqrt{2}$ | $\omega^{2}A/\sqrt{2}$ | $E/2$ | $E/2$ |
Phase relations. Velocity is a quarter of a cycle ahead of displacement (phase lead $\pi/2$). Acceleration is exactly opposite to displacement (phase difference $\pi$). All three have the same period T. When the particle crosses the mean position, x is zero but v is at its peak, so you can see why they are a quarter cycle apart.
Worked example: reading an SHM equation. A particle moves as $x = 0.06\cos(5\pi t + \pi/3)$, with x in metres and t in seconds. Find the amplitude, period, frequency, initial displacement, maximum speed and maximum acceleration.
| Step | What we do | Result |
|---|---|---|
| 1 | Compare with $x = A\cos(\omega t + \phi)$ | $A = 0.06$ m, $\omega = 5\pi$ rad s⁻¹, $\phi = \pi/3$ |
| 2 | Period from $\omega = 2\pi/T$ | $T = 2\pi/5\pi = 0.4$ s |
| 3 | Frequency $\nu = 1/T$ | $\nu = 2.5$ Hz |
| 4 | Put t = 0 | $x = 0.06\cos(\pi/3) = 0.03$ m |
| 5 | $v_{\max} = A\omega$ | $0.06 \times 5\pi = 0.3\pi \approx 0.94$ m s⁻¹ |
| 6 | $a_{\max} = A\omega^{2}$ | $0.06 \times 25\pi^{2} = 1.5\pi^{2} \approx 14.8$ m s⁻² |
The common wrong step: calling 5$\pi$ the frequency. The number multiplying t is the angular frequency $\omega$. The frequency is $\omega/2\pi = 2.5$ Hz.
Pause and predict: a particle in SHM of amplitude A has speed half of its maximum. How far is it from the mean position?
Check your answer: $v = \omega\sqrt{A^{2} - x^{2}} = A\omega/2$ gives $\sqrt{A^{2} - x^{2}} = A/2$, so $x^{2} = 3A^{2}/4$ and $x = (\sqrt{3}/2)A$, about 0.87 A. The particle is closer to the end than to the middle, because it is already slowing down there.
Worked example: velocity and acceleration at a given point. A particle has amplitude 5 cm and $\omega = 10$ rad s⁻¹. Find its speed and acceleration when it is 3 cm from the mean position.
| Step | What we do | Result |
|---|---|---|
| 1 | Use $v = \omega\sqrt{A^{2} - x^{2}}$, not $A\omega$, because it is not at the mean position | $v = 10 \times \sqrt{25 - 9}$ cm s⁻¹ |
| 2 | $\sqrt{16} = 4$ | v = 40 cm s⁻¹ = 0.40 m s⁻¹ |
| 3 | Acceleration magnitude $\omega^{2}x$ | 100 × 3 = 300 cm s⁻² = 3 m s⁻² |
| 4 | Direction | Towards the mean position |
The common wrong step: using $v_{\max} = A\omega$ = 50 cm s⁻¹ at every point, or writing $v = \omega x$. The speed depends on where the particle is.
Timing in a cycle. If the particle starts at the mean position, $x = A\sin\omega t$, and it reaches A/2 after T/12 and the end after T/4. If it starts at an extreme, $x = A\cos\omega t$, and it reaches A/2 after T/6 and the mean position after T/4. Going from one extreme to the other always takes T/2.
Force and energy: why the motion is simple
By Newton's second law, $F = ma = -m\omega^{2}x$. Writing $k = m\omega^{2}$, the force is $F = -kx$: a restoring force proportional to displacement and directed towards the mean position. Any system with this force is a linear harmonic oscillator, and its angular frequency is
$$\omega = \sqrt{k/m}$$
This is the physics behind the formula. A stiffer system (large k) pulls back harder, so it oscillates faster. A heavier system (large m) is harder to accelerate, so it oscillates slower. Forces with extra $x^{2}$ or $x^{3}$ terms give non-linear oscillators, which are not SHM.
Energy. Kinetic energy is $K = (1/2)mv^{2} = (1/2)m\omega^{2}(A^{2} - x^{2}) = (1/2)k(A^{2} - x^{2})$. Potential energy stored in the restoring force is $U = (1/2)kx^{2}$. Adding them,
$$E = K + U = \dfrac{1}{2}kA^{2} = \dfrac{1}{2}m\omega^{2}A^{2}$$
The total never changes. Energy simply flows back and forth: all kinetic at the mean position, all potential at the ends. As functions of time, $K = (1/2)kA^{2}\sin^{2}(\omega t + \phi)$ and $U = (1/2)kA^{2}\cos^{2}(\omega t + \phi)$. A squared sine repeats twice as fast as a sine, so K and U each have period T/2, while x, v and a have period T. Because $E \propto A^{2}$, doubling the amplitude makes the energy four times larger, while the period stays the same.
Pause and predict: you stretch a spring-block system twice as far before releasing it. What happens to the period, the maximum speed and the total energy?
Check your answer: the period does not change (it depends only on m and k). The maximum speed doubles, since $v_{\max} = A\omega$. The energy becomes four times larger, since $E = (1/2)kA^{2}$. A bigger stretch means a bigger force and a longer path, and the two effects cancel in the period.
Worked example: energy of a spring-block system. A 0.5 kg block on a smooth horizontal surface is attached to a spring of k = 200 N m⁻¹. It is pulled 10 cm and released. Find the total energy, $\omega$, $v_{\max}$, and the kinetic and potential energy when it is 6 cm from the mean position.
| Step | What we do | Result |
|---|---|---|
| 1 | $E = (1/2)kA^{2}$ with A = 0.10 m | $(1/2) \times 200 \times 0.01 = 1.0$ J |
| 2 | $\omega = \sqrt{k/m}$ | $\sqrt{200/0.5} = 20$ rad s⁻¹ |
| 3 | $v_{\max} = A\omega$, then confirm with $(1/2)mv^{2}$ | 2 m s⁻¹, and $(1/2) \times 0.5 \times 4 = 1$ J |
| 4 | At x = 0.06 m, $U = (1/2)kx^{2}$ | $(1/2) \times 200 \times 0.0036 = 0.36$ J |
| 5 | $K = E - U$ | 1.0 − 0.36 = 0.64 J |
| 6 | Same answer from $v = \omega\sqrt{A^{2} - x^{2}} = 20 \times 0.08 = 1.6$ m s⁻¹ | $(1/2) \times 0.5 \times 2.56 = 0.64$ J |
The common wrong step: taking the energy at x = 6 cm as $(1/2)kx^{2} = 0.36$ J. That is only the potential part. The total energy of the oscillator is fixed at 1.0 J by the amplitude.
Beyond the textbook. For the exam: total energy stays constant, $E = (1/2)kA^{2}$. In reality: every real oscillator loses energy to friction and air drag, so its amplitude slowly shrinks (damping). The period barely changes for light damping, but the energy is no longer constant. If you push the system periodically at a frequency close to its natural frequency $\omega = \sqrt{k/m}$, the amplitude can grow very large. That is resonance, the reason a swing goes high when you push in time with it. Why it matters: CUET asks the ideal, undamped case, but it helps to know that "constant energy" is an idealisation.
Spring-mass systems
For a block on a spring, $T = 2\pi\sqrt{m/k}$. For a block hanging from a spring, gravity only moves the equilibrium position down, to where the spring is stretched by $e = mg/k$. The oscillation about that new position is still SHM with the same k and m, so gravity does not enter the period. You can also write the period as $T = 2\pi\sqrt{e/g}$, where e is the static extension of the spring under the load.
Combining springs. Think of what each arrangement does to the force on the block.
| Arrangement | Effective spring constant | Reason |
|---|---|---|
| Parallel (both springs stretch by the same x) | $k_{1} + k_{2}$ | The forces add |
| Two springs on opposite sides of a block | $k_{1} + k_{2}$ (so $2k$ for identical springs) | When one stretches the other compresses, and both push the block back |
| Series (end to end) | $k_{1}k_{2}/(k_{1} + k_{2})$ | The extensions add, the force is the same; softer overall |
| Spring cut into n equal parts | Each piece has $k \times n$ | A shorter spring is stiffer; k times length is constant |
Worked example: a hanging mass and springs in series. (a) A mass hung on a spring stretches it by 9.8 cm at equilibrium. Find the period of small vertical oscillations (g = 9.8 m s⁻²). (b) A 2 kg block is attached to two springs of 300 N m⁻¹ and 600 N m⁻¹ joined end to end. Find the period.
| Step | What we do | Result |
|---|---|---|
| (a) 1 | Use the static extension: $T = 2\pi\sqrt{e/g}$ | $2\pi\sqrt{0.098/9.8}$ |
| (a) 2 | $\sqrt{0.01} = 0.1$ | $T = 0.2\pi \approx 0.63$ s |
| (b) 1 | Series combination | $k = (300 \times 600)/(300 + 600) = 200$ N m⁻¹ |
| (b) 2 | $T = 2\pi\sqrt{m/k}$ | $2\pi\sqrt{2/200} = 2\pi/10 \approx 0.63$ s |
The common wrong step: for (b), adding the constants to get 900 N m⁻¹. Springs end to end are softer than either one alone, so k must come out smaller than 300.
Beyond the textbook. For the exam: the spring is massless. In reality: a real spring has its own mass $m_{s}$, and part of it moves with the block. A good approximation is to replace $m$ by $m + m_{s}/3$ in the period formula. Why it matters: it explains why a measured period from a lab experiment is always a little longer than $2\pi\sqrt{m/k}$.
The simple pendulum
A small heavy bob on a light, inextensible string of length L is pulled aside by an angle $\theta$ and released. Gravity mg splits into a part along the string, balanced by the tension, and a part mg sin $\theta$ along the arc that pulls the bob back. The restoring torque about the support is $-mgL\sin\theta$, and with moment of inertia $mL^{2}$ this gives
$$\alpha = -(g/L)\sin\theta$$
For small angles $\sin\theta \approx \theta$ (in radians), so $\alpha = -(g/L)\theta$. That is SHM in the angle, with $\omega = \sqrt{g/L}$ and
$$T = 2\pi\sqrt{L/g}$$
Why does the mass not appear? The restoring force $mg\sin\theta$ is proportional to m, and the inertia that resists it is also proportional to m. They cancel. The period also does not depend on amplitude, as long as the angle is small. The approximation $\sin\theta \approx \theta$ is very good for small angles: at 10° the error is about 0.5%, and at 20° about 2%.
| Situation | Effective g | Period of the same pendulum |
|---|---|---|
| Normal surface | g | $T_{0} = 2\pi\sqrt{L/g}$ |
| Moon ($g/6$) | $g/6$ | $\sqrt{6} \approx 2.45$ times longer |
| Length increased by 21% | g | $\sqrt{1.21} = 1.1$ times longer |
A pendulum that "ticks seconds" has $T = 2$ s, so $L = gT^{2}/4\pi^{2} = 9.8 \times 4/39.5 \approx 0.99$ m, about 1 m.
Worked example: a pendulum on the Moon. A seconds pendulum ($T = 2.0$ s) on the Earth is taken to the Moon, where g is one sixth. (a) Find its period there. (b) What length would keep a 2.0 s period on the Moon?
| Step | What we do | Result |
|---|---|---|
| (a) 1 | $T \propto 1/\sqrt{g}$, so $T_{\text{Moon}} = T_{\text{Earth}} \times \sqrt{g_{\text{Earth}}/g_{\text{Moon}}}$ | $2.0 \times \sqrt{6}$ |
| (a) 2 | $\sqrt{6} \approx 2.45$ | $T \approx 4.9$ s |
| (b) 1 | Keep T fixed, so $L/g$ must be fixed. $L_{\text{Moon}} = L_{\text{Earth}} \times (1/6)$ | $0.99 / 6$ |
| (b) 2 | Result | $L \approx 0.17$ m |
The common wrong step: concluding that the pendulum swings faster because gravity is weaker. A weaker pull means a slower return, so the period gets longer.
Pause and predict: a pendulum clock is accurate in a city at sea level. It is taken to a mountain top where g is very slightly smaller. Does it run fast or slow?
Check your answer: slow. Smaller g gives a longer period, so each tick takes more time than the clock thinks it does and it falls behind.
Beyond the textbook. For the exam: a pendulum is SHM only for small angles, and $T = 2\pi\sqrt{L/g}$ is independent of amplitude. In reality: for larger swings the period grows. At an amplitude of 60° it is about 7% longer than the small-angle value, and it keeps rising towards infinity as the amplitude approaches 180°. Galileo's observation that a pendulum keeps time regardless of amplitude is therefore an excellent approximation, not an exact law. Why it matters: it shows what "valid for small angles" really costs you, and why old clocks used a small swing.
Exam traps
| The trap | Why it's tempting | The truth |
|---|---|---|
| "Every periodic motion is SHM." | Both repeat | SHM must be sinusoidal, with $a = -\omega^{2}x$. A bouncing ball and the Earth's orbit are periodic but not SHM |
| "Speed is maximum at the extremes." | The extremes look like the fastest part of a swing | Speed is maximum at the mean position (A$\omega$) and zero at the ends. Acceleration is the reverse |
| "Kinetic energy has period T." | x has period T | K and U each repeat every T/2, because they depend on squares |
| "Doubling the amplitude doubles the energy." | Energy seems to grow with size | $E = (1/2)kA^{2}$, so doubling A gives four times E |
| "Two identical springs on either side give $T = 2\pi\sqrt{m/k}$." | Only one formula is memorised | Both springs act, so k becomes 2k and $T = 2\pi\sqrt{m/2k}$ |
| "The pendulum period depends on the bob's mass or the amplitude." | Heavier or larger sounds like slower | $T = 2\pi\sqrt{L/g}$ has neither, for small angles |
| "$\omega$ is the frequency." | Both are called frequency | $\omega = 2\pi\nu$. Read $x = A\cos(8\pi t)$ as $\nu = 4$ Hz, not $8\pi$ Hz |
| "A pendulum runs faster on the Moon." | Weaker gravity sounds like less drag | $T \propto 1/\sqrt{g}$. Weaker g means a longer period |
Formula sheet
| Quantity | Formula | SI unit | When to use |
|---|---|---|---|
| Period and frequency | T = 1/$\nu$, $\omega = 2\pi/T = 2\pi\nu$ | s, Hz, rad s⁻¹ | Converting any time information |
| Displacement | $x = A\cos(\omega t + \phi)$ | m | Position at time t |
| Velocity | v = −A$\omega$ sin($\omega t + \phi$) | m s⁻¹ | Speed at time t |
| Velocity at position x | $v = \pm \omega\sqrt{A^{2} - x^{2}}$ | m s⁻¹ | Speed when x is given |
| Maximum speed | $v_{\max} = A\omega$ (at x = 0) | m s⁻¹ | Mean position |
| Acceleration | $a = -\omega^{2}x$ | m s⁻² | Any displacement |
| Maximum acceleration | $a_{\max} = \omega^{2}A$ (at x = ±A) | m s⁻² | Extreme positions |
| Restoring force | F = −kx, $k = m\omega^{2}$ | N | Dynamics of the oscillator |
| Spring period | $T = 2\pi\sqrt{m/k}$ | s | Block-spring problems |
| Spring in series and parallel | $k = k_{1}k_{2}/(k_{1} + k_{2})$ and $k = k_{1} + k_{2}$ | N m⁻¹ | Combined springs |
| Kinetic energy | $K = (1/2)k(A^{2} - x^{2})$ | J | Energy at a position |
| Potential energy | $U = (1/2)kx^{2}$ | J | Energy at a position |
| Total energy | $E = (1/2)kA^{2} = (1/2)m\omega^{2}A^{2}$ | J | Constant for the whole motion |
| Pendulum period | $T = 2\pi\sqrt{L/g}$ | s | Small angles only |
60-second revision
| Topic | Lock this in |
|---|---|
| Definition | SHM means $x = A\cos(\omega t + \phi)$, equivalently $a = -\omega^{2}x$ |
| Reference circle | SHM is the projection of uniform circular motion on a diameter |
| Where things peak | v max and a = 0 at the mean. a max and v = 0 at the ends |
| Phase | v leads x by $\pi$/2. a is opposite to x (difference $\pi$) |
| Speed at x | $v = \omega\sqrt{A^{2} - x^{2}}$ |
| Energy | $E = (1/2)kA^{2}$, K and U period T/2, $K = U$ at $x = A/\sqrt{2}$ |
| Spring | $T = 2\pi\sqrt{m/k}$. Gravity does not change it. Opposite springs add, series springs soften |
| Pendulum | $T = 2\pi\sqrt{L/g}$. No mass, no amplitude (small angles). Longer on the Moon |
| Timings | Mean to A/2: T/12. Extreme to A/2: T/6. Quarter cycle: T/4 |
Where this chapter connects
- Before this: Laws of Motion gives F = ma, and Work, Energy and Power gives the energy exchange you use for $E = (1/2)kA^{2}$.
- Same idea elsewhere: Mechanical Properties of Solids explains why a spring obeys F = −kx (Hooke's law), and Gravitation explains why g, and so the pendulum period, changes with place.
- Next: Waves is what you get when many oscillators are coupled, and Alternating Current brings back the same sine curves for voltage and current.
🎯 Practice MCQs
First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed
Q1. Which of the following statements is correct?
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Answer: B
Oscillatory motion repeats, so it is always periodic. Uniform circular motion is periodic but does not go to and fro about a mean position, so the converse fails. (A) is false because SHM needs a sinusoidal form. (C) reverses the truth. (D) is meaningless, since SHM is periodic for any amplitude.
Q2. A particle moves as $x = 0.04\cos(8\pi t + \pi/3)$, where x is in metres and t in seconds. Its frequency and phase constant are respectively
▸ Show answer & explanation
Answer: B
Comparing with $A\cos(\omega t + \phi)$ gives $\omega = 8\pi$ rad per second and $\phi = \pi/3$. The frequency is $\nu = \omega/2\pi = 4$ Hz. (A) treats $\omega$ as the frequency. (C) mixes up the phase constant. (D) forgets to divide by 2$\pi$ properly.
Q3. A particle in SHM has amplitude 4 cm and period $\pi/5$ s. Its maximum speed is
▸ Show answer & explanation
Answer: A
$\omega = 2\pi/T = 2\pi/(\pi/5) = 10$ rad per second. $v_{\max} = A\omega = 0.04 \times 10 = 0.4$ m per second. (B) is just A. (C) forgets to convert 4 cm to metres. (D) multiplies A by T instead of by $\omega$.
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Q4. A particle in SHM has $\omega = 2$ rad s⁻¹. When its displacement is $x = -0.1$ m, its acceleration is
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Answer: B
$a = -\omega^{2}x = -4 \times (-0.1) = +0.4$ m per second squared. The positive sign means it points towards the mean position, since the particle is on the negative side. (A) has the wrong sign and direction. (C) uses $\omega$ instead of $\omega^{2}$. (D) holds only at $x = 0$.
Q5. A block of mass m on a spring of constant k = 100 N m⁻¹ oscillates with amplitude 0.2 m. If the amplitude is doubled, the new total energy compared with the old one is
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Answer: C
$E = (1/2)kA^{2}$. The old energy is $(1/2) \times 100 \times 0.04 = 2$ J and the new one is $(1/2) \times 100 \times 0.16 = 8$ J, which is four times. (B) assumes $E$ is proportional to $A$ rather than $A^{2}$. (A) would be true for the period, not for the energy.
Q6. A block of mass 2 kg is attached to a spring of constant 200 N m⁻¹ on a frictionless surface. The period of its oscillation is
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Answer: A
$T = 2\pi\sqrt{m/k} = 2\pi\sqrt{2/200} = 2\pi \times (1/10) = \pi/5$ s. (B) comes from forgetting the square root and writing $2\pi \times m/k \times 10$. (C) halves the correct answer. (D) takes $\sqrt{m/k}$ as $1/2$.
Q7. A simple pendulum that ticks seconds (period 2 s) at a place where g = 9.8 m s⁻² has a length of about
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Answer: C
$L = gT^{2}/4\pi^{2} = 9.8 \times 4/39.5 \approx 0.99$ m. (D) results from forgetting the $4\pi^{2}$ factor and using L = T. (A) and (B) come from using T/2 as the period.
Q8. A 1 kg block on a spring of constant 50 N m⁻¹ is pulled 10 cm from the mean position and released from rest. When the block is at 5 cm, the ratio of kinetic to potential energy is
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Answer: B
$E = (1/2) \times 50 \times 0.01 = 0.25$ J. U at 5 cm = $(1/2) \times 50 \times 0.0025 = 0.0625$ J, so K = 0.25 − 0.0625 = 0.1875 J. K : U = 0.1875 : 0.0625 = 3 : 1. (A) is the inverse. (C) holds only at $x = A/\sqrt{2}$, about 7.1 cm.
Q9. For a particle in SHM, the phase difference of velocity with respect to displacement, and of acceleration with respect to displacement, are respectively
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Answer: C
$v = -A\omega \sin(\omega t + \phi) = A\omega \cos(\omega t + \phi + \pi/2)$ is a quarter cycle ahead of $x$. $a = -\omega^{2}x$ is exactly opposite to $x$, a difference of $\pi$. (D) wrongly says v is in phase with x. (B) swaps the two.
Q10. A block of mass m lies on a smooth surface between two identical springs, each of constant k, attached on opposite sides to fixed walls. The period of small oscillations is
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Answer: B
If the block moves by x, one spring stretches and the other compresses, and both push it back. The net force is $-2kx$, so the effective constant is $2k$ and $T = 2\pi\sqrt{m/2k}$. (A) counts only one spring. (C) puts the 2 in the wrong place. (D) is a distractor with no basis.
Q11. Assertion (A): In SHM the kinetic energy and the potential energy each have a period equal to half the period of the displacement. Reason (R): Kinetic and potential energies depend on the squares of $\sin(\omega t + \phi)$ and $\cos(\omega t + \phi)$, which repeat twice in each cycle of x.
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Answer: A
$K = (1/2)kA^{2}\sin^{2}(\omega t + \phi)$ and $U = (1/2)kA^{2}\cos^{2}(\omega t + \phi)$. A squared sine or cosine repeats every half period, which is exactly what the assertion states. So R explains A.
Q12. Match List I with List II. List I: (P) Particle at the mean position (Q) Particle at an extreme position (R) Particle at $x = A/\sqrt{2}$ (S) Particle at $x = A/2$. List II: (1) Speed is maximum and acceleration is zero (2) Acceleration is maximum and total energy is entirely potential (3) Kinetic energy equals potential energy (4) Kinetic energy is three fourths of the total energy
▸ Show answer & explanation
Answer: A
At $x = 0$ the speed is $A\omega$ and $a = 0$, so P-1. At x = ±A the speed is zero, a is maximum and E = U, so Q-2. $K = U$ needs $(1/2)k(A^{2} - x^{2}) = (1/2)kx^{2}$, giving $x = A/\sqrt{2}$, so R-3. At $x = A/2$, $U = E/4$, so $K = 3E/4$, giving S-4. (C) swaps R and S.
Q13. A simple pendulum has a period of 2.0 s on the Earth. Its period on a planet where g is one sixth of the Earth value is approximately
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Answer: C
$T \propto 1/\sqrt{g}$, so $T = 2.0 \times \sqrt{6} = 2.0 \times 2.45 \approx 4.9$ s. (A) divides by $\sqrt{6}$ instead of multiplying. (D) multiplies by 6, forgetting the square root. (B) wrongly assumes T does not depend on g.
Q14. A particle in SHM has amplitude 5 cm and angular frequency 10 rad s⁻¹. Its maximum acceleration is
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Answer: B
$a_{\max} = \omega^{2}A = 100 \times 0.05 = 5$ m per second squared. (A) uses $\omega A = 0.5$, which is the maximum speed in m per second. (C) and (D) forget to convert 5 cm to metres.
Q15. A mass hung from a vertical spring stretches it by 9.8 cm at equilibrium. It is pulled down slightly and released. Taking g = 9.8 m s⁻², the period of its oscillations is
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Answer: B
At equilibrium $kx = mg$, so $m/k = e/g = 0.098/9.8 = 0.01$ s². $T = 2\pi\sqrt{m/k} = 2\pi \times 0.1 = 0.63$ s. (A) forgets the $2\pi$. (C) and (D) take the pendulum value for a 1 m length, which has no connection with this spring.
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