Exams › JEE Advanced › Physics
The shape of a string transmitting a transverse wave along the x-axis at a certain instant is shown in a figure. The transverse velocity of point P is v = 4*pi cm/s and the slope angle at P satisfies theta = arctan(0.004*pi). Which of the following statements are correct?
- Amplitude of the wave is 2 mm.
- Speed of wave propagation is 10 m/s.
- Maximum acceleration of a particle on the string is 80*pi² cm/s².
- The wave is traveling in the negative x-direction.
Correct answer: Speed of wave propagation is 10 m/s.
Solution
The wave speed equals the ratio of the transverse particle speed to the magnitude of the string slope at that point: v_wave = v_P / |tan(theta)| = 4*pi / (0.004*pi) = 1000 cm/s = 10 m/s.
Related JEE Advanced Physics questions
⚔️ Practice JEE Advanced Physics free + battle 1v1 →