A stretched string fixed at both ends has m n | vedclass

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

Question

(A) For a stretched string fixed at both ends,the length $l$ is given by $l = \frac{p \lambda}{2}$,where $p$ is the number of loops (or segments). In

Vedclass · Accepted Answer

$(m-1) \frac{\lambda}{2}$

Vedclass · Answer

(A) For a stretched string fixed at both ends,the length $l$ is given by $l = \frac{p \lambda}{2}$,where $p$ is the number of loops (or segments). In a standing wave on a string fixed at both ends,the number of loops $p$ is equal to the number of anti-nodes. The relationship between the number of nodes $m$ and the number of anti-nodes $p$ is given by $m = p + 1$. Therefore,the number of loops is $p = m - 1$. Substituting this value of $p$ into the length formula,we get $l = \frac{(m-1) \lambda}{2}$.

$A$ stretched string fixed at both ends has $m$ nodes,then the length of the string will be

Similar Questions

Two travelling waves ${y_1} = A\sin [k(x - ct)]$ and ${y_2} = A\sin [k(x + ct)]$ are superimposed on a string. The distance between adjacent nodes is

At a certain instant,a stationary transverse wave is found to have maximum kinetic energy. The appearance of the string at that instant is:

For a stationary wave,$Y = 10 \sin \left( \frac{\pi x}{15} \right) \cos (48 \pi t) \text{ cm}$,the distance between a node and the successive antinode is (in $\text{ cm}$)

In stationary waves,all particles between two consecutive nodes pass through the mean position:

In a stationary wave,which of the following statements is correct regarding strain?

Vedclass Test Series

Exam Paper Generator

Online Exam Module