Question

The displacement of a particle executing SHM is \[ x=3\sin2t+4\cos2t. \] The amplitude of particle is

• A. \(7\)
• B. \(3\)
• C. \(4\)
• D. \(5\)

Hint:

For \(a\sin\omega t+b\cos\omega t\), amplitude is \(\sqrt{a^2+b^2}\).

Answer 1

The Correct Option is D

Concept:
If displacement is of the form: \[ x=a\sin\omega t+b\cos\omega t \] then amplitude is: \[ A=\sqrt{a^2+b^2} \]

Step 1: Given: \[ x=3\sin2t+4\cos2t \]

Step 2: Compare with: \[ x=a\sin\omega t+b\cos\omega t \] \[ a=3,\qquad b=4 \]

Step 3: Amplitude: \[ A=\sqrt{3^2+4^2} \] \[ A=\sqrt{9+16} \] \[ A=\sqrt{25}=5 \] \[ \boxed{5} \]