Resultant Amplitude of A sin wt + B cos wt

Physics · Oscillations · NEET

When you add A sin wt and B cos wt, the result is still simple harmonic motion with the SAME frequency. Its resultant amplitude is R = square root of (A squared + B squared), and the phase constant is phi = tan inverse (B/A). Memory hook: sine and cosine are like the two legs of a right triangle, so the amplitude is the hypotenuse, just like Pythagoras theorem.
A sin wt + B cos wt combine like a right triangleBAR = root(A squared + B squared)phitan phi = B / APeaks at different timesso add at right anglesNot A + B
A sin wt and B cos wt act like two sides at right angles. The resultant amplitude R is the hypotenuse, root(A squared + B squared), and the phase phi satisfies tan phi = B/A. The plain sum A + B is wrong.

Your doubts, answered

Is A sin wt + B cos wt actually SHM?

Yes. Both terms have the same angular frequency w. Their sum can be rewritten as a single term R sin(wt + phi). Any motion of the form R sin(wt + phi) is SHM, so the sum is SHM with the same frequency w and time period T = 2 pi / w. Only the amplitude and starting phase change, not the frequency.

Why is the amplitude root(A squared + B squared) and not A + B?

Because sin wt and cos wt reach their maximum values at different times. Sine is maximum when cosine is zero, and cosine is maximum when sine is zero. So the two peaks never add up at the same moment. Think of them as two sides at right angles: you combine them like a right triangle, giving the hypotenuse R = root(A squared + B squared), never the plain sum A + B.

How do I find the phase constant phi?

Write A sin wt + B cos wt = R sin(wt + phi). Expand the right side: R sin wt cos phi + R cos wt sin phi. Match terms: A = R cos phi and B = R sin phi. Divide them: tan phi = B/A, so phi = tan inverse (B/A). Squaring and adding gives R = root(A squared + B squared).

Does the constant A0 in y = A0 + A sin wt + B cos wt change the amplitude?

No. A0 is a fixed number added to every position. It only shifts the mean (centre) position of the oscillation up or down. The particle still swings the same distance about this new centre, so the amplitude stays R = root(A squared + B squared). A0 is not part of the amplitude.

What if the two terms have different frequencies, like A sin w1 t + B cos w2 t?

Then the simple formula does NOT apply. R = root(A squared + B squared) works ONLY when both terms share the SAME angular frequency w. With different frequencies the motion is not simple harmonic and has no single fixed amplitude.

⚠️ The NEET trap
Amplitude = A + B (just add the two coefficients).
Amplitude = root(A squared + B squared), because sine and cosine peak at different instants and combine at right angles.
🧠 For 3 sin wt + 4 cos wt the trap answer is 7, but the correct amplitude is root(9 + 16) = root 25 = 5.

Real NEET questions

2019

The displacement of a particle executing simple harmonic motion is y = A0 + A sin wt + B cos wt. The amplitude of its oscillation is:

A · A0 + root(A squared + B squared)
B · root(A squared + B squared)
C · root(A0 squared + (A + B) squared)
D · A + B
Solution: Step 1: A0 is a constant offset. It only shifts the mean position and does not add to the amplitude, so ignore it. Step 2: The oscillating part is A sin wt + B cos wt. Both have the same frequency w, so write it as R sin(wt + phi). Step 3: Matching gives A = R cos phi and B = R sin phi. Square and add: R squared = A squared + B squared. Step 4: So amplitude R = root(A squared + B squared). Correct option is B.

Solved Oscillations NEET PYQs

Try the real previous-year questions from this chapter — each with the answer and a full solution.

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Frequently asked

What is the resultant amplitude of A sin wt + B cos wt?

It is R = root(A squared + B squared). This single amplitude describes the combined simple harmonic motion, which has the same frequency as the two parts.

What is the phase constant of A sin wt + B cos wt?

The phase constant is phi = tan inverse (B/A), found by writing the sum as R sin(wt + phi) and matching the sine and cosine coefficients.

Is the sum A sin wt + B cos wt periodic and simple harmonic?

Yes, as long as both terms have the same angular frequency w. The sum equals R sin(wt + phi), which is SHM with time period T = 2 pi / w.

What is the amplitude of 3 sin wt + 4 cos wt?

It is root(3 squared + 4 squared) = root(9 + 16) = root 25 = 5 units. The common trap answer 7 (from 3 + 4) is wrong.

Can I also write the sum using a cosine instead of a sine?

Yes. You can write A sin wt + B cos wt = R cos(wt - alpha) with the same R = root(A squared + B squared). Only the phase reference changes; the amplitude is identical.