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SHM Systems
Defining SHM
Simple harmonic motion (SHM) is oscillatory motion where the restoring force is directly proportional to displacement from equilibrium and always directed towards it.
$$a = -\omega^2 x$$
Acceleration is proportional to displacement and opposite in direction — the defining signature of SHM.
$a$=Instantaneous acceleration(m/s²)
$\omega$=[Angular Frequency](rad/s)
$x$=Displacement from equilibrium(m)
$x = 0$ (equilibrium)
→Acceleration is zero, velocity is maximum.
$x = \pm A$ (extremes)
→Acceleration is maximum ($|a_{max}| = \omega^2 A$), velocity is zero.
Key Test: If $a \propto -x$, the motion is SHM. The negative sign means acceleration always points back to the mean position.
Amplitude ($A$): Maximum displacement from equilibrium — does NOT affect period or frequency in ideal SHM.
Dimensional Check: $[\omega^2 x] = (\text{rad/s})^2 \times \text{m} = \text{m/s}^2$ ✓ confirms acceleration units.
SHM Quantities
SHM is characterized by angular frequency $\omega$, time period $T$, and frequency $f$, all interrelated through simple formulas.
$$\omega = \frac{2\pi}{T} = 2\pi f$$
Angular frequency converts between the time domain (period) and the circular motion analogy (radians per second).
$\omega$=Angular frequency(rad/s)
$T$=Time for one complete oscillation(s)
$f$=Number of oscillations per second(Hz)
Vibration: One complete round trip — mean → extreme → opposite extreme → back to mean.
Unit Check: $[2\pi / T] = 1/\text{s} = \text{rad/s}$ since radians are dimensionless.
SHM and Circular Motion
SHM is the projection of uniform circular motion onto a diameter — a point moving in a circle at constant $\omega$ produces SHM when viewed from the side.
Displacement: $x = A \sin(\omega t)$ — the projection of the rotating point onto the diameter.
Velocity: $v = A\omega \cos(\omega t) = \omega\sqrt{A^2 - x^2}$ — maximum at center ($x = 0$), zero at extremes ($x = \pm A$).
Acceleration: $a = -A\omega^2 \sin(\omega t) = -\omega^2 x$ — always opposite to displacement.
Proportionality Shortcut: $v_{max} = A\omega$ and $a_{max} = A\omega^2$. If $\omega$ doubles, $v_{max}$ doubles but $a_{max}$ quadruples.
Phase of Oscillation
The phase angle $\theta = \omega t + \phi$ specifies both the displacement and direction of motion at any instant.
$$x = A \sin(\omega t + \phi)$$
The general SHM equation with initial phase $\phi$ — determines where in the cycle the motion begins.
$A$=Amplitude(m)
$\omega t$=Phase gained over time(rad)
$\phi$=Initial phase at $t = 0$(rad)
$\phi = 0$
→Starts from mean position: $x = A \sin(\omega t)$.
$\phi = \pi/2$
→Starts from positive extreme: $x = A \cos(\omega t)$.
Phase Determines State: At phase $0$ the body is at mean position moving forward; at $\pi/2$ it is at the positive extreme; at $\pi$ back at mean moving backward; at $3\pi/2$ at the negative extreme.
Phase Difference: Two oscillators with a phase difference of $\pi$ are always on opposite sides of equilibrium (anti-phase).
The Mass-Spring System
The mass-spring system is a fundamental oscillator where a restoring force proportional to displacement (Hooke's Law) drives motion.
$$T = 2\pi \sqrt{\frac{m}{k}}$$
The period depends on the ratio of inertia to stiffness.
$T$=Time period(s)
$m$=Oscillating mass(kg)
$k$=[Spring Constant](N/m)
Increasing $m$ by factor $n$
→$T$ increases by $\sqrt{n}$ — e.g. quadrupling mass doubles $T$.
Increasing $k$ by factor $n$
→$T$ decreases by $\sqrt{n}$ — stiffer spring means faster oscillation.
Vertical spring
→Same formula applies — gravity only shifts the equilibrium point downward, not the period.
Proportionality: $T \propto \sqrt{m}$ and $T \propto 1/\sqrt{k}$. These square-root relationships are a common source of errors.
Equilibrium Position: The point where net force is zero ($x = 0$); velocity is maximum here.
Restoring Force: Always directed towards equilibrium ($F = -kx$), opposing the displacement.
Finding $k$ from Static Extension: Hang a known mass and measure extension: $k = mg/x$. Use this $k$ with a different mass to find $T$.
Dimensional Check: $[2\pi\sqrt{m/k}] = \sqrt{\text{kg}/(\text{N/m})} = \sqrt{\text{s}^2} = \text{s}$ ✓
The instantaneous velocity of a mass on a spring depends on both position and the system's angular frequency.
$$v = \sqrt{\frac{k}{m}(A^2 - x^2)}$$
Velocity is maximum at center and zero at the extremes — it depends on how far the mass is from its turning point.
$v$=Instantaneous speed(m/s)
$A$=Amplitude(m)
$x$=Current displacement(m)
$x = 0$
→$v_{max} = A\sqrt{k/m} = A\omega$.
$x = \pm A$
→$v = 0$ — the mass momentarily stops at the extremes.
Maximum Speed: $v_{max} = A\omega = A\sqrt{k/m}$. Doubling amplitude doubles $v_{max}$.
Normalized Form: $v = v_{max}\sqrt{1 - x^2/A^2}$ — useful for finding speed at any fraction of amplitude.
The Simple Pendulum
A simple pendulum approximates SHM for small angles, where gravity acts as the restoring agent.
$$T = 2\pi \sqrt{\frac{L}{g}}$$
The period depends solely on geometry and local gravity, independent of mass.
$L$=Effective length of string(m)
$g$=[Gravitational Field Strength](m/s²)
$\theta > 15°$
→The Small Angle Approximation fails; motion is no longer strictly SHM.
$L \to \infty$
→$T \to \infty$ — an infinitely long pendulum would never complete a swing.
$g \to 0$ (free fall / space)
→$T \to \infty$ — no gravity means no restoring force, so no oscillation.
Small Angle Approximation: Valid when $\sin \theta \approx \theta$ in radians, ensuring force is linear with displacement.
Mass Independence: Increasing the bob's mass increases both inertia and weight, which cancel out in the period calculation.
Restoring Force Derivation: The tangential component $F = -mg\sin\theta \approx -mg\theta = -mg(x/L)$, giving $a = -(g/L)x$, confirming SHM with $\omega = \sqrt{g/L}$.
Proportionality: $T \propto \sqrt{L}$ and $T \propto 1/\sqrt{g}$. Quadrupling $L$ doubles $T$; taking the pendulum to a planet with $g/4$ also doubles $T$.
Energy in SHM
SHM systems continuously swap energy between kinetic and potential forms while maintaining a constant Mechanical Energy.
$$E_{total} = \frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$$
Total energy is proportional to the square of the amplitude.
$E_{total}$=Total mechanical energy(J)
$A$=Amplitude (maximum displacement)(m)
$v_{max}$=Maximum velocity at equilibrium(m/s)
$x = A/2$
→$PE = E/4$ and $KE = 3E/4$ — at half-amplitude, 75% of energy is kinetic.
$KE = PE$
→This occurs at $x = A/\sqrt{2} \approx 0.707A$.
Turning Points: At maximum displacement, velocity is zero and all energy is potential (elastic or gravitational).
Maximum Velocity: Occurs at the equilibrium point ($x=0$) where all potential energy has been converted to kinetic.
Energy at Any Point: $KE = \frac{1}{2}k(A^2 - x^2)$ and $PE = \frac{1}{2}kx^2$. Their sum is always $\frac{1}{2}kA^2$.
Proportionality Shortcut: $E \propto A^2$. Doubling amplitude quadruples total energy.
Application Strategy
Worked Examples
Problem Statement
A block weighing 4.0 kg extends a spring by 0.16 m from its unstretched position. The block is removed and a 0.50 kg body is hung from the same spring. What is its period of vibration?
Spring Constant from Static Extension$T = 2\pi\sqrt{m/k}$
Strategic Toolkit
$k = \frac{mg}{x}, \quad T = 2\pi \sqrt{\frac{m}{k}}$
Step 1
Find Spring Constant
Use the static extension of the 4.0 kg block: $k = F/x = mg/x$.
$k = \frac{4.0 \times 9.8}{0.16} = \frac{39.2}{0.16} = 245$ N/m
Step 2
Apply Period Formula
Use $k = 245$ N/m with the new mass $m = 0.50$ kg.
$T = 2\pi \sqrt{\frac{0.50}{245}} = 2\pi \sqrt{0.00204} = 2\pi(0.0452) \approx 0.28$ s
Problem Statement
What should be the length of a simple pendulum whose period is 1.0 s at a place where $g = 9.8$ m/s²?
Pendulum Period$T = 2\pi\sqrt{L/g}$
Strategic Toolkit
$L = \frac{gT^2}{4\pi^2}$
Step 1
Rearrange Formula
Square both sides and solve for $L$.
$T^2 = 4\pi^2 \frac{L}{g} \Rightarrow L = \frac{gT^2}{4\pi^2}$
Step 2
Calculation
Substitute $T = 1.0$ s and $g = 9.8$ m/s².
$L = \frac{9.8 \times 1.0^2}{4 \times (3.14)^2} = \frac{9.8}{39.5} \approx 0.25$ m
Step 3
Find Frequency
Frequency is the reciprocal of period.
$f = 1/T = 1/1.0 = 1$ Hz
Problem Statement
A 0.4 kg mass on a spring ($k = 100$ N/m) oscillates with amplitude 0.2 m. Find the speed when the mass is 0.1 m from equilibrium.
SHM Velocity Formula$v = \omega\sqrt{A^2 - x^2}$
Strategic Toolkit
$v = \sqrt{\frac{k}{m}(A^2 - x^2)}$
Step 1
Find Angular Frequency
Calculate $\omega$ from the spring and mass.
$\omega = \sqrt{k/m} = \sqrt{100/0.4} = \sqrt{250} \approx 15.8$ rad/s
Step 2
Apply Velocity Formula
Substitute $A = 0.2$ m and $x = 0.1$ m.
$v = 15.8\sqrt{0.2^2 - 0.1^2} = 15.8\sqrt{0.03} = 15.8 \times 0.173 \approx 2.74$ m/s
Memorize formulas
Mass-Spring Period
Inertia ($m$) on top, stiffness ($k$) on bottom.
Formula
$T = 2\pi \sqrt{\frac{m}{k}}$
Pendulum Period
Length ($L$) on top, gravity ($g$) on bottom. Mass-independent.
Formula
$T = 2\pi \sqrt{\frac{L}{g}}$
SHM Acceleration
Proportional to displacement, opposite in direction. The defining equation of SHM.
Formula
$a = -\omega^2 x$
SHM Velocity
Maximum at center, zero at extremes. Depends on how far from the turning point.
Formula
$v = \omega\sqrt{A^2 - x^2}$
SHM Displacement
General equation with initial phase $\phi$.
Formula
$x = A\sin(\omega t + \phi)$
Review
1
A mass $m$ on a single spring of constant $k$ oscillates with period $T$. The spring is replaced by two identical springs, each of constant $k$, connected in parallel, and the mass is tripled to $3m$. What is the new period?
Ans
$T\sqrt{3/2}$
Review
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Find the new spring constant
Two identical springs in parallel add their constants, so the effective stiffness doubles.
Substitute into the SHM period formula
Use $T = 2\pi\sqrt{m/k}$ with the new mass and effective spring constant, then compare with the original period.
$$T' = 2\pi\sqrt{\frac{3m}{2k}} = T\sqrt{\frac{3}{2}}$$
2
A simple pendulum of length $L$ has a period $T$ on Earth. If taken to a planet where gravity is $g/4$, what length must the pendulum be to maintain the same period $T$?
Ans
$L/4$
Review
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Match the pendulum ratio
For a simple pendulum, the period depends on $\sqrt{L/g}$. To keep the same period, the quantity $L/g$ must stay unchanged.
Scale length with gravity
If gravity becomes $g/4$, the length must also become one quarter so the ratio stays the same.
$$\frac{L'}{g/4} = \frac{L}{g} \Rightarrow L' = \frac{L}{4}$$
3
In a vertical mass-spring system, where is the magnitude of acceleration at its maximum?
Ans
At the maximum displacement points
Review
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Use the SHM acceleration law
In SHM, acceleration is proportional to displacement and opposite in direction, so its magnitude grows with $|x|$.
Check the extreme positions
The displacement is largest at the turning points, so the acceleration magnitude is also largest there.
$$a = -\omega^2 x \quad \Rightarrow \quad |a| \text{ is maximum when } |x| \text{ is maximum}$$
4
Which change increases the frequency ($f$) of a simple pendulum?
Ans
Shortening the string
Review
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Read the frequency formula
A pendulum's frequency depends on $\sqrt{g/L}$, so the only listed change that increases $f$ is making $L$ smaller.
Eliminate the other choices
Mass and amplitude do not affect the ideal pendulum frequency, and lowering $g$ reduces frequency instead of increasing it.
$$f = \frac{1}{2\pi}\sqrt{\frac{g}{L}} \Rightarrow L \downarrow \; \Rightarrow \; f \uparrow$$
5
A particle in SHM has total energy $E$. At what displacement from the mean position is the kinetic energy equal to the potential energy?
Ans
$A/\sqrt{2}$
Review
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Split the total energy equally
If kinetic and potential energy are equal, each must be half of the total energy.
Use the SHM potential energy expression
Write $PE = \frac{1}{2}kx^2$ and $E = \frac{1}{2}kA^2$, then set $PE = E/2$ and solve for $x$.
$$\frac{1}{2}kx^2 = \frac{1}{2}E = \frac{1}{4}kA^2 \Rightarrow x = \frac{A}{\sqrt{2}}$$
6
A body executing SHM starts from rest at $x = +A$ at $t = 0$. Which equation describes its displacement?
Ans
$x = A\cos(\omega t)$
Review
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Match the initial position
The displacement must be $+A$ at $t=0$, so the function should start at its maximum value.
Check the initial velocity
Because the body starts from rest, the derivative at $t=0$ must be zero. The cosine form satisfies both conditions directly.
$$x = A\cos(\omega t), \quad x(0)=A, \quad v(0)=\frac{dx}{dt}\bigg|_{t=0}=0$$
7
A 3.0 kg block stretches a vertical spring by 0.12 m. The block is removed and a 0.75 kg body is hung from the same spring and set oscillating. What is the period of oscillation?
Ans
0.35 s
Review
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Find the spring constant from static stretch
At equilibrium, the weight equals the spring force, so $kx = mg$ for the 3.0 kg block.
Use the mass-spring period formula
Replace the block with the 0.75 kg oscillator and substitute the spring constant into $T = 2\pi\sqrt{m/k}$.
$$k = \frac{mg}{x} = \frac{(3.0)(9.8)}{0.12} = 245\ \text{N/m}$$
$$T = 2\pi\sqrt{\frac{0.75}{245}} \approx 0.35\ \text{s}$$
8
The maximum velocity of a mass on a spring is $v_{max}$. If the amplitude is doubled while keeping the same spring and mass, the new maximum velocity is:
Ans
$2v_{max}$
Review
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Use the maximum-speed relation
For SHM on a spring, the peak speed is proportional to amplitude through $v_{max} = A\omega$.
Double the amplitude
Since the angular frequency is unchanged, doubling $A$ doubles the maximum speed.
$$v'_{max} = (2A)\omega = 2v_{max}$$
9
A particle executes SHM with angular frequency $\omega$. At displacement $x = A/2$, its speed is:
Ans
$\frac{\sqrt{3}}{2} A\omega$
Review
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Apply the SHM speed formula
The speed at displacement $x$ is given by $v = \omega\sqrt{A^2 - x^2}$.
Substitute the given displacement
Put in $x = A/2$ and simplify the square root.
$$v = \omega\sqrt{A^2 - \left(\frac{A}{2}\right)^2} = \omega\sqrt{\frac{3A^2}{4}} = \frac{\sqrt{3}}{2}A\omega$$
10
A point P moves uniformly on a circle of radius $r$ with angular speed $\omega$. The projection of P on any diameter executes SHM. If P completes one full revolution, the projection has completed:
Ans
Exactly one complete oscillation
Review
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Connect circular motion to SHM
The projection follows a sine or cosine function of the angle, so one full angular cycle gives one full SHM cycle.
Count the cycles
A $2\pi$ revolution is exactly one period of the projection's motion, so it completes one oscillation.
$$2\pi \text{ revolution} \rightarrow 1 \text{ SHM cycle}$$
Question 1 of 10
1
A mass $m$ on a single spring of constant $k$ oscillates with period $T$. The spring is replaced by two identical springs, each of constant $k$, connected in parallel, and the mass is tripled to $3m$. What is the new period?