For a periodic motion represented by the equation
y=sin ω t+cos ω t
the amplitude of the motion is

Solution

Physics sir · 3 steps

Sine and cosine are two SHMs a quarter-cycle apart, so their amplitudes add like perpendicular vectors.

1. Rewrite as one sine. Take √2 out as a common factor: y = √2(1/√2 sin ω t + 1/√2 cos ω t). Since cos 45° = sin 45° = 1/√2, the bracket is sin(ω t + 45°).

   y = √2 sin(ω t + π/4)

2. Read the amplitude. This is a single SHM with angular frequency ω. Its amplitude is the number in front of the sine, which is √2. The same result comes from A = √(a² + b²) = √(1² + 1²).
3. So. The amplitude of the motion is √2, option (b).

The idea to keep. Any a sin ω t + b cos ω t is one SHM with amplitude √(a² + b²), because the two parts are 90° out of phase. Never add their amplitudes directly.

Answer from the NTA final answer key. Solution by PrepWiser.