Question

If the displacement \( y \) (in cm) of a particle executing simple harmonic motion is given by the equation: \[ y = 5 \sin(3 \pi t) + 5 \sqrt{3} \cos(3 \pi t) \] then the amplitude of the particle is:

• A. \( 5 \, \text{cm} \)
• B. \( 5 \sqrt{3} \, \text{cm} \)
• C. \( 5 (1 + \sqrt{3}) \, \text{cm} \)
• D. \( 10 \, \text{cm} \)

Hint:

For any equation of form \( y = a\sin \theta + b\cos \theta \), the amplitude is always \( \sqrt{a^2 + b^2} \).

Answer 1

The Correct Option is D

Step 1: Rewrite in standard form
The given equation is: \[ y = 5 \sin(3 \pi t) + 5 \sqrt{3} \cos(3 \pi t) \] Step 2: Use trigonometric identity
We can express this as: \[ y = A \sin(3 \pi t + \phi) \] where amplitude \( A = \sqrt{a^2 + b^2} \) Step 3: Calculate amplitude
Here \( a = 5 \), \( b = 5 \sqrt{3} \): \[ A = \sqrt{5^2 + (5 \sqrt{3})^2} = \sqrt{25 + 75} = \sqrt{100} = 10 \, \text{cm} \]