$A$ string fixed at one end and free at the other is vibrating in its second overtone. The length of the string is $10 \ cm$ and the maximum amplitude of vibration of particles of the string is $2 \ mm$. Then the amplitude of the particle at $9 \ cm$ from the fixed end is

  • A
    $\sqrt{3} \ mm$
  • B
    $\sqrt{2} \ mm$
  • C
    $\frac{\sqrt{3}}{2} \ mm$
  • D
    None of these

Explore More

Similar Questions

The figure shows a stationary wave between two fixed points $P$ and $Q$. Which point$(s)$ among $1, 2,$ and $3$ are in phase with point $X$?

$A$ wave is represented by the equation $y = 10 \sin 2\pi(100t - 0.02x) + 10 \sin 2\pi(100t + 0.02x)$. The maximum amplitude and loop length are respectively:

Difficult
View Solution

In a standing wave on a string rigidly fixed at both ends:

Difficult
View Solution

$A$ standing wave pattern is formed on a string. One of the waves is given by the equation $y_1 = a \cos(\omega t - kx + \pi/3)$. Find the equation of the other wave such that at $x = 0$,a node is formed.

The equation of a stationary wave is $y = 10 \sin \left( \frac{\pi x}{4} \right) \cos (20 \pi t)$. The distance between two consecutive nodes is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo