JEE Mains Physics · Oscillations

Timing in SHM and Combining SHMs

The time between two positions is the phase covered divided by ω; two SHMs of the same frequency on one line add like vectors into one SHM.

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Why this matters

Seventeen PYQs, eight of them asking for a number, and three from 2026. Ten ask for the time to go between two positions, or work back from such a time to the period. Seven add two SHMs or ask whether a given function is simple harmonic at all.

Concept 1 of 2: Time taken to move between two positions in SHM

The phase grows at a steady rate ω, but the particle does not move at a steady speed. So time is found from the phase, never from the distance. Write the motion with a sine if it starts at the mean position and with a cosine if it starts at an extreme; then read the phase at each position.

Definition

From the mean: $x = A\sin\omega t$ . From an extreme: $x = A\cos\omega t$ .
Time $= \dfrac{\text{phase covered}}{\omega} = \dfrac{\text{phase covered}}{2\pi}\,T$ .
Mean to $A/2$ : $T/12$ . $A/2$ to $A$ : $T/6$ . Mean to $A/\sqrt{2}$ : $T/8$ . $A$ to $A/\sqrt{2}$ : $T/8$ . $A$ to $\sqrt{3}A/2$ : $T/12$ . Mean to extreme: $T/4$ .
Distance: $4A$ in each full period and $2A$ in each half period. A quarter period adds exactly A only when it starts at the mean or at an extreme.
Energy instants (from the mean, $U \propto \sin^{2}\omega t$ ): K = U first at $T/8$ ; U is largest at $T/4$ ; the slope $dU/dt \propto \sin 2\omega t$ is largest first at $T/8$ .

Time from the phase covered

t = \frac{\Delta\theta}{\omega} = \frac{\Delta\theta}{2\pi}\,T

Worked example

A particle does SHM with period 12 s and amplitude 4 cm. Find (a) the time to go from

x = A

x = A/2

, (b) the time to go from the mean position to

x = A/\sqrt{2}

, and (c) the distance it covers in 15 s starting from the mean position.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q12Moderate

Example 1 · Oscillations · Timing in SHM and Combining SHMs

A particle is executing simple harmonic motion. Its amplitude is A and time period is 5 sec . the time required by it to move from

x=A

x=\frac{A}{\sqrt{2}}

\_\_\_\_

sec.

Starting at an extreme needs the cosine

From x = A to x = A/2 is a phase of π/3 in x = A cos ωt, so it takes T/6. Reading it with x = A sin ωt gives T/12, which is the time from the mean to A/2, a different journey.

Equal distances do not take equal times

Mean to A/2 takes T/12, but A/2 to A takes T/6, twice as long, because the particle slows down near the extreme.

A quarter period covers A only from the mean or an extreme

Starting anywhere else, the distance in T/4 is not A. In a half period the distance is always 2A, wherever the motion starts.

Drill 9 more on time taken to move between two positions in shm

Concept 2 of 2: Adding two SHMs of the same frequency, and telling SHM from periodic motion

Two SHMs along one line with the same ω add like two vectors (phasors) at an angle equal to their phase difference. The result is one SHM with the same ω. A motion is SHM only if it can be written as one sine of ωt, perhaps shifted by a constant; a sum of different frequencies may repeat, but it is not SHM.

Definition

Same ω, phase difference $\Delta\phi$ : $A = \sqrt{A_1^{2} + A_2^{2} + 2A_1A_2\cos\Delta\phi}$ .
In phase: $A_1 + A_2$ . Opposite in phase: $|A_1 - A_2|$ . At $\pi/2$ : $\sqrt{A_1^{2} + A_2^{2}}$ .
$a\sin\omega t + b\cos\omega t = \sqrt{a^{2} + b^{2}}\,\sin(\omega t + \phi)$ with $\tan\phi = b/a$ .
$\sin^{2}\omega t = \dfrac{1}{2} - \dfrac{1}{2}\cos 2\omega t$ : SHM about $x = \dfrac{1}{2}$ with period $\pi/\omega$ .
$\sin^{3}\omega t = \dfrac{3\sin\omega t - \sin 3\omega t}{4}$ : periodic, not SHM.
$\cos\omega t + \cos 2\omega t$ : periodic with period $2\pi/\omega$ , not SHM.
$\sin\omega t + \cos\pi\omega t$ : the two periods are in the ratio π, which is irrational, so the motion never repeats.

Resultant of two SHMs of the same frequency

A = \sqrt{A_1^{2} + A_2^{2} + 2A_1A_2\cos\Delta\phi} \qquad a\sin\omega t + b\cos\omega t = \sqrt{a^{2} + b^{2}}\,\sin\left(\omega t + \tan^{-1}\frac{b}{a}\right)

Worked example

A particle is subjected to

x_1 = 6\sin 4t

cm and

x_2 = 6\sin\left(4t + \dfrac{2\pi}{3}\right)

cm at the same time. Find the amplitude of the resulting motion and its maximum acceleration.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 2 Apr 2025 · Q9Moderate

Example 2 · Oscillations · Timing in SHM and Combining SHMs

A particle is subjected two simple harmonic motions as :

x_{1}=\sqrt{7}\sin 5t\ cm

and

x_{2}= 2\sqrt{7}\sin\left( 5t+\frac{\pi}{3} \right)cm

where x is displacement and

t

is time in seconds. The maximum acceleration of the particle is

x \times10^{- 2}{\text{ }ms}^{- 2}

. The value of x is :

Amplitudes add as vectors, not as numbers

Amplitudes 3 and 4 at a phase difference of π/2 give 5, not 7. Only SHMs exactly in phase give A₁ + A₂.

A constant shift does not spoil SHM

sin²ωt equals 1/2 − (1/2)cos 2ωt. That is SHM about x = 1/2, with angular frequency 2ω and period π/ω, half the period of sin ωt.

Two different frequencies never make SHM

cos ωt + cos 2ωt repeats every 2π/ω but is not SHM. If the two periods have an irrational ratio, as in sin ωt + cos πωt, the motion does not repeat at all.

Drill 6 more on adding two shms of the same frequency, and telling shm from periodic motion

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Time taken to move between two positions in SHM
Time from the phase covered
$t = \frac{\Delta\theta}{\omega} = \frac{\Delta\theta}{2\pi}\,T$
Adding two SHMs of the same frequency, and telling SHM from periodic motion
Resultant of two SHMs of the same frequency
$A = \sqrt{A_1^{2} + A_2^{2} + 2A_1A_2\cos\Delta\phi} \qquad a\sin\omega t + b\cos\omega t = \sqrt{a^{2} + b^{2}}\,\sin\left(\omega t + \tan^{-1}\frac{b}{a}\right)$

Watch out for (6)

Starting at an extreme needs the cosine
→ Time taken to move between two positions in SHM
Equal distances do not take equal times
→ Time taken to move between two positions in SHM
A quarter period covers A only from the mean or an extreme
→ Time taken to move between two positions in SHM
Amplitudes add as vectors, not as numbers
→ Adding two SHMs of the same frequency, and telling SHM from periodic motion
A constant shift does not spoil SHM
→ Adding two SHMs of the same frequency, and telling SHM from periodic motion
Two different frequencies never make SHM
→ Adding two SHMs of the same frequency, and telling SHM from periodic motion

Test yourself on Oscillations

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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