Question

The displacement of a particle executing SHM is \(x = 3 \sin 2t + 4 \cos 2t\). The amplitude of particle is

• A. 7
• B. 3
• C. 4
• D. 5

Hint:

This is a direct application of the superposition of two perpendicular vectors (or phasors in this context). If you have a vector with components 3 and 4, its magnitude is 5. This is a classic 3-4-5 Pythagorean triple, which often appears in physics problems. Recognizing it can provide an instant answer.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given an equation for the displacement of a particle which is a sum of a sine and a cosine function with the same frequency. We need to find the amplitude of the resulting Simple Harmonic Motion (SHM).

Step 2: Key Formula or Approach:
An expression of the form \(x = a \sin(\omega t) + b \cos(\omega t)\) represents an SHM. The amplitude \(A\) of this resultant motion is given by:
\[ A = \sqrt{a^2 + b^2} \]
The expression can be rewritten as \(x = A \sin(\omega t + \phi)\) or \(x = A \cos(\omega t + \delta)\).

Step 3: Detailed Explanation:
The given equation for displacement is:
\[ x = 3 \sin(2t) + 4 \cos(2t) \]
This matches the standard form \(x = a \sin(\omega t) + b \cos(\omega t)\) with:
- \(a = 3\)
- \(b = 4\)
- \(\omega = 2\) rad/s
Now, we can calculate the resultant amplitude \(A\) using the formula:
\[ A = \sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2} \]
\[ A = \sqrt{9 + 16} = \sqrt{25} \]
\[ A = 5 \]
The unit of the amplitude would be the same as the unit of displacement \(x\).

Step 4: Final Answer:
The amplitude of the particle is 5.