Question

The instantaneous displacement of a particle executing simple harmonic motion is given by \(x = 2(\cos(\pi t) + \sin(\pi t))\). 
The amplitude of oscillation is

• A. \(3\sqrt{2}\)
• B. 4
• C. \(4\sqrt{2}\)
• D. \(2\sqrt{2}\)
• E. \(8\sqrt{2}\)

Hint:

Using trigonometric identities can simplify the analysis of oscillatory motion and provide insights into the physical quantities involved, like amplitude.

Answer 1

The Correct Option is D

To solve the problem, we analyze the given displacement equation for simple harmonic motion (SHM):

\[ x = 2(\cos(\pi t) + \sin(\pi t)) \] The general form of the displacement equation for SHM is: \[ x = A\cos(\omega t) + B\sin(\omega t) \] where:
\(A\) and \(B\) are constants,
\(\omega\) is the angular frequency.
The amplitude \(R\) of the oscillation is given by: \[ R = \sqrt{A^2 + B^2} \] 
Step 1: Identify \(A\), \(B\), and \(\omega\) From the given equation: \[ x = 2(\cos(\pi t) + \sin(\pi t)) \] we can rewrite it as: \[ x = 2\cos(\pi t) + 2\sin(\pi t) \] Thus:
\(A = 2\),
\(B = 2\),
\(\omega = \pi\).
Step 2: Calculate the amplitude Using the formula for amplitude: \[ R = \sqrt{A^2 + B^2} \] Substitute \(A = 2\) and \(B = 2\): \[ R = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \] Final Answer: The amplitude of oscillation is: \[ \boxed{2\sqrt{2}} \]