For a periodic motion represented by the equation y = t + t, the amplitude of the motion is:

Given: y = t + t Find: The amplitude of the motion. For the form y = A ( t) + B ( t) the amplitude is A^2 + B^2 Here, A = 1 and B = 1. So, Amplitude = (1)^2 + (1)^2 = 2 Therefore, the amplitude is 2. Hence, the correct option is D. The solution states option A, but the worked result clearly gives 2, which matches option D.

For a periodic motion represented by the equation y = t + t, the… — Solution | JEE Physics

MCQEasyJEE 2023Simple Harmonic Motion (SHM)

For a periodic motion represented by the equation $y = \sin \omega t + \cos \omega t$ , the amplitude of the motion is:

A
$0.5$
B
$1$
C
$2$
D
$\sqrt{2}$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: $y = \sin \omega t + \cos \omega t$

Find: The amplitude of the motion.

For the form

y = A \sin(\omega t) + B \cos(\omega t)

the amplitude is

\sqrt{A^2 + B^2}

Here, $A = 1$ and $B = 1$ .

So,

\text{Amplitude} = \sqrt{(1)^2 + (1)^2} = \sqrt{2}

Therefore, the amplitude is $\sqrt{2}$ . Hence, the correct option is D.

The solution states option A, but the worked result clearly gives $\sqrt{2}$ , which matches option D.

Common mistakes

Taking the amplitude as $A + B = 2$ is incorrect because the resultant amplitude for $A \sin(\omega t) + B \cos(\omega t)$ is not the arithmetic sum. Use $\sqrt{A^2 + B^2}$ instead.
Assuming the amplitude is $1$ by looking at the coefficients separately is wrong because both sine and cosine terms contribute together to the resultant SHM. Combine them using the standard formula.

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Answer

Step-by-step solution

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