$A$ stretched string is fixed at both ends. It is made to vibrate so that the total number of nodes formed in it is '$x$'. The length of the string in terms of the wavelength of waves formed in it is $(\lambda = \text{wavelength})$
- A$\frac{x \lambda}{2}$
- B$\left(x+\frac{1}{2}\right) \frac{\lambda}{2}$
- C$(x-1) \frac{\lambda}{2}$
- D$(x+1) \frac{\lambda}{2}$
Explore More
Similar Questions
$A$ string fixed at both the ends forms a standing wave with a node separation of $5 \,cm$. If the velocity of the wave on the string is $2 \,m/s$, then the frequency of vibration of the string is (in $\,Hz$)
Stationary waves are set up in an air column. The velocity of sound in air is $330 \ m/s$ and the frequency is $165 \ Hz$. The distance between the nodes is ... $m$.
Easy
View Solution$A$ standing wave is formed by the superposition of two waves travelling in opposite directions. The transverse displacement is given by $y(x, t) = 0.5 \sin(\frac{5\pi}{4}x) \cos(200\pi t)$. What is the speed of the travelling wave moving in the positive $x$ direction in $m/s$? ($x$ and $t$ are in meter and second,respectively.)
DifficultJEE MAIN 2017
View SolutionLength of a string tied to two rigid supports is $40 \ cm$. The maximum wavelength (in $cm$) of a stationary wave produced on it is ... $cm$.
$A$ stationary wave is represented by $y = 10 \sin \left( \frac{\pi x}{4} \right) \cos (20 \pi t)$,where $x$ and $y$ are in $cm$ and $t$ is in seconds. The distance between two consecutive nodes is (in $cm$)
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo