JEE Main SHM PYQ

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6. (JEE Main 2024 (Online) 29th January Evening Shift)

A simple harmonic oscillator has an amplitude $A$ and time period $6 π$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x =$ A to $x = \frac{\sqrt{3}}{2}$ A will be $\frac{π}{x} s$ , where $x =$ _________.

Correct answer is 2

JEE Main 2024 (Online) 29th January Evening Shift Physics - Simple Harmonic Motion Question 8 English Explanation

From phasor diagram particle has to move from $P$ to $Q$ in a circle of radius equal to amplitude of SHM.

$\begin{aligned} \cos ϕ = \frac{\frac{\sqrt{3} A}{2}}{A} = \frac{\sqrt{3}}{2} \\ ϕ = \frac{π}{6} \end{aligned}$

Now, $\frac{π}{6} = ω t$

$\begin{aligned} \frac{π}{6} = \frac{2 π}{T} t \\ \frac{π}{6} = \frac{2 π}{6 π} t \end{aligned}$

$t = \frac{π}{2}$

So, $x = 2$

7. (JEE Main 2024 (Online) 27th January Morning Shift)

A particle executes simple harmonic motion with an amplitude of $4 cm$ . At the mean position, velocity of the particle is $10 cm / s$ . The distance of the particle from the mean position when its speed becomes $5 cm / s$ is $\sqrt{α} cm$ , where $α =$ ________.

Correct answer is 12

$\begin{aligned} V_{at mean position} = A ω \Rightarrow 10 = 4 ω \\ ω = \frac{5}{2} \\ V = ω \sqrt{A^{2} - x^{2}} \\ 5 = \frac{5}{2} \sqrt{4^{2} - x^{2}} \Rightarrow x^{2} = 16 - 4 \\ x = \sqrt{12} cm \end{aligned}$