A particle is subjected to two simple harmonic motions as : x1 = √7 sin 5t cm and x2 = 2√7 sin(5t + π/3) cm where x is displacement and t is time in seconds. The maximum acceleration of the particle is x x 10-2 ms-2. The value of x is :

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q33 – JEE-MAIN 2025

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Home Physics OscillationsQuestion #392

A particle is subjected to two simple harmonic motions as :

x_{1} = \sqrt{7} \sin 5 t

cm and

x_{2} = 2 \sqrt{7} \sin (5 t + \frac{π}{3})

cm where

x

is displacement and

t

is time in seconds. The maximum acceleration of the particle is

x \times 10^{- 2}

^{- 2}

. The value of

x

is :

Clue 1 of 3: Superposition of SHM

When two simple harmonic motions with the same angular frequency are superimposed, the resultant motion is also a simple harmonic motion.

🥷

Ninja Strategy•Factor of Angular Frequency Squared

Recognize that maximum acceleration in SHM is $A ω^{2}$ . Since $ω = 5$ , $ω^{2} = 25$ , so the answer must be a multiple of 25, which eliminates option 2.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Identify Parameters and Phase Difference✦ Active

The given simple harmonic motions are $x_{1} = \sqrt{7} \sin 5 t$ cm and $x_{2} = 2 \sqrt{7} \sin (5 t + \frac{π}{3})$ cm. Comparing these with the general form $x = A \sin (ω t + δ)$ , we identify the amplitudes, angular frequency, and phases:

A_{1} = \sqrt{7} cm, A_{2} = 2 \sqrt{7} cm

ω = 5 rad/s

δ_{1} = 0, δ_{2} = \frac{π}{3}

The phase difference is ϕ = δ_{2} - δ_{1} = \frac{π}{3}

Step 2: Calculate Resultant Amplitude○ Expand

The amplitude of the resultant simple harmonic motion ( $A_{r e s}$ ) due to the superposition of two SHMs with the same angular frequency is given by:

A_{r e s} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}

A_{r e s} = \sqrt{(\sqrt{7})^{2} + (2 \sqrt{7})^{2} + 2 (\sqrt{7}) (2 \sqrt{7}) \cos (\frac{π}{3})}

A_{r e s} = \sqrt{7 + (4 \times 7) + (4 \times 7 \times \frac{1}{2})}

A_{r e s} = \sqrt{7 + 28 + 14} = \sqrt{49}

A_{r e s} = 7 cm

💡 Teacher's Secret Hint

Remember that $\cos (\frac{π}{3}) = \frac{1}{2}$ .

Step 3: Determine Maximum Acceleration and Final Value○ Expand

The maximum acceleration ( $a_{m a x}$ ) of a particle in SHM is given by $a_{m a x} = A_{r e s} ω^{2}$ . We have $A_{r e s} = 7$ cm and $ω = 5$ rad/s.

a_{m a x} = 7 cm \times (5 rad/s)^{2}

a_{m a x} = 7 \times 25 = 175 {cm/s}^{2}

The question asks for the value of $x$ where $a_{m a x} = x \times 10^{- 2}$ ms $^{- 2}$ . First, convert $a_{m a x}$ to m/s $^{2}$ :

175 {cm/s}^{2} = 175 \times 10^{- 2} {m/s}^{2}

Comparing this with $x \times 10^{- 2}$ ms $^{- 2}$ , we find $x = 175$ .

💡 Teacher's Secret Hint

Pay attention to units and the format requested for the final answer.

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Physics Question 33 – JEE-MAIN 2025

Clue 1 of 3: Superposition of SHM

🧠 Full Solution Path