Physics Q37 – JEE-MAIN 2026

A particle is executing simple harmonic motion. Its amplitude is

A

and time period is

5

sec. The time required by it to move from

x = A

x = \frac{A}{\sqrt{2}}

is _______ sec.

Clue 1 of 3: Understanding SHM

Recall the general equation for displacement in Simple Harmonic Motion and how to determine the phase constant based on initial conditions.

🥷

Ninja Strategy•Standard SHM Time Fractions

Recognize that the time to reach $x = A / \sqrt{2}$ from $x = A$ is a standard fraction of the time period, specifically $T / 8$ . With $T = 5$ sec, this directly gives $5 / 8$ sec.

🧠 Full Solution Path

Step 1: Set up the SHM equation✦ Active

Since the particle starts from $x = A$ (maximum positive displacement) at $t = 0$ , the appropriate displacement equation for Simple Harmonic Motion is:

x (t) = A \cos (ω t)

Step 2: Determine the time for the given displacement○ Expand

We need to find the time $t$ when the displacement is $x = \frac{A}{\sqrt{2}}$ . Substitute this into the equation:

\frac{A}{\sqrt{2}} = A \cos (ω t)

\cos (ω t) = \frac{1}{\sqrt{2}}

ω t = \frac{π}{4}

💡 Teacher's Secret Hint

Remember that $\cos (θ) = \frac{1}{\sqrt{2}}$ for $θ = \frac{π}{4}$ (in the first quadrant).

Step 3: Calculate the time using the given period○ Expand

The angular frequency $ω$ is related to the time period $T$ by $ω = \frac{2 π}{T}$ . Substitute this into the equation from Step 2:

\frac{2 π}{T} t = \frac{π}{4}

Given $T = 5$ sec, we can solve for $t$ :

\frac{2 π}{5} t = \frac{π}{4}

t = \frac{π}{4} \times \frac{5}{2 π}

t = \frac{5}{8} sec

💡 Teacher's Secret Hint

Ensure units are consistent throughout the calculation.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 37 – JEE-MAIN 2026

Clue 1 of 3: Understanding SHM

🧠 Full Solution Path