A string fixed at one end and free at the oth | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For a string fixed at one end and free at the other,the length $L$ is given by $L = (2n+1) \frac{\lambda}{4}$,where $n$ is the overtone number.For

Vedclass · Accepted Answer

$\sqrt{2} \ mm$

Vedclass · Answer

(B) For a string fixed at one end and free at the other,the length $L$ is given by $L = (2n+1) \frac{\lambda}{4}$,where $n$ is the overtone number.For the second overtone,$n = 2$,so $L = 5 \frac{\lambda}{4}$.Given $L = 10 \ cm$,we have $10 = \frac{5 \lambda}{4}$,which gives $\lambda = 8 \ cm$.The amplitude $A(x)$ at a distance $x$ from the fixed end is given by $A(x) = A_{max} \sin(kx)$,where $k = \frac{2\pi}{\lambda}$.Here $A_{max} = 2 \ mm$ and $x = 9 \ cm$.$A(9) = 2 \sin\left(\frac{2\pi}{8} 	imes 9ight) = 2 \sin\left(\frac{9\pi}{4}ight) = 2 \sin\left(2\pi + \frac{\pi}{4}ight) = 2 \sin\left(\frac{\pi}{4}ight)$.$A(9) = 2 	imes \frac{1}{\sqrt{2}} = \sqrt{2} \ mm$.

$A$ 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

Similar Questions

$A$ stationary wave is represented by $y = 12 \cos \left(\frac{\pi}{6} x\right) \sin (8 \pi t)$,where $x$ and $y$ are in $cm$ and $t$ is in seconds. The distance between two successive antinodes is (in $cm$)

$A$ wave $y = a \sin(\omega t - kx)$ on a string meets with another wave producing a node at $x = 0$. Then the equation of the unknown wave is

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 $X$?

$A$ stationary wave is represented by $y = A \sin(100t) \cos(0.01x)$,where $y$ and $A$ are in millimetres,$t$ is in seconds,and $x$ is in metres. The velocity of the constituent wave is ........... $m/s$.

The equation $\overrightarrow{\phi}(x, t) = \overrightarrow{j} \sin \left( \frac{2\pi}{\lambda} vt \right) \cos \left( \frac{2\pi}{\lambda} x \right)$ represents:

Vedclass Test Series

Exam Paper Generator

Online Exam Module