Displacement of a wave is expressed as x(t)=5cos(628t + pi/2) m. The wavelength of the wave when its velocity is 300 m/s is : (pi=3.14)

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q34 – JEE-MAIN 2025

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Displacement of a wave is expressed as

x (t) = 5 \cos (628 t + \frac{π}{2})

m. The wavelength of the wave when its velocity is

300

m/s is : (

π = 3.14

)

Clue 1 of 3: Wave Equation Analysis

Compare the given wave equation with the standard form to identify key parameters like angular frequency.

🧠 Full Solution Path

Step 1: Identify Angular Frequency✦ Active

The given displacement equation for a wave is $x (t) = 5 \cos (628 t + \frac{π}{2})$ m. Comparing this to the standard wave equation $x (t) = A \cos (ω t + ϕ)$ , we can identify the angular frequency $ω$ .

ω = 628 rad/s

Step 2: Calculate Frequency○ Expand

The relationship between angular frequency ( $ω$ ) and frequency ( $f$ ) is $ω = 2 π f$ . We can use this to find the frequency $f$ . Use the given value of $π = 3.14$ .

f = \frac{ω}{2 π} = \frac{628}{2 \times 3.14} = \frac{628}{6.28} = 100 Hz

Step 3: Calculate Wavelength○ Expand

The relationship between wave velocity ( $v$ ), frequency ( $f$ ), and wavelength ( $λ$ ) is $v = f λ$ . We are given the velocity $v = 300$ m/s and have calculated the frequency $f = 100$ Hz. We can now find the wavelength $λ$ .

λ = \frac{v}{f} = \frac{300 m/s}{100 Hz} = 3 m

💡 Teacher's Secret Hint

Ensure units are consistent throughout the calculation.

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Physics Question 34 – JEE-MAIN 2025

Clue 1 of 3: Wave Equation Analysis

🧠 Full Solution Path