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JEE Main 20257 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Evening Shift
Subject
Physics
Chapter
Waves and Sound

JEE Main 2025 · 7 April · Evening Shift · Physics · Waves and Sound

The equation of a wave travelling on a string is y= [20 π x+10 π t]...

The equation of a wave travelling on a string is y=sin⁡[20πx+10πt]\mathrm{y}=\sin [20 \pi \mathrm{x}+10 \pi \mathrm{t}], where x and t are distance and time in SI units. The minimum distance between two points having the same oscillating speed is :

Options

  1. A5.0 cm
  2. B20 cm
  3. C10 cm
  4. D2.5 cm

Correct answer

A. 5.0 cm

Step-by-step solution

Minimum distance between 2 points having same speed is λ2\frac{\lambda}{2}.λ=2πk=110m=10cm Distance =λ2=\lambda=\frac{2 \pi}{\mathrm{k}}=\frac{1}{10} \mathrm{ m}=10 \mathrm{ cm} \text{ Distance }= \frac{\lambda}{2}=5 cm\mathrm{ cm}
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