Question

The displacement of a particle is given by $x = 3 \sin(5\pi t) + 4 \cos(5\pi t)$. The amplitude of particle is:

• A. 3
• B. 4
• C. 5
• D. 7

Hint:

The phase difference between sine and cosine functions of the same frequency is $90^{\circ}$.

Answer 1

The Correct Option is C

Step 1: Concept
For a superposition of two perpendicular SHMs $x = a \sin(\omega t) + b \cos(\omega t)$, the resultant amplitude is $A = \sqrt{a^{2} + b^{2}}$.
Step 2: Analysis
Here $a = 3$ and $b = 4$. $A = \sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16}$.
Step 3: Conclusion
$A = \sqrt{25} = 5$.
Final Answer: (C)