Name: Interference of Waves on a String
Uploaded: 2019-07-19
Duration: 6 min 24 s
Description: Wave interference is the phenomenon that occurs when two waves meet while traveling along the same medium.Here we will study the interference of waves in string. Let's watch.

Question

A light string is tied at one end to fixed support and to a heavy string of equal length $L$ at the other end as shown in figure. Mass per unit length of the strings are $μ$ and $9 μ$ and the tension is $T .$ Find the possible values of frequencies such that point $A$ is a node/antinode.

Accepted Answer

Let us refer to the lighter and heavier strings as strings $1$ and $2$ , respectively.

Speed of sound in a string is given by,

$v = \sqrt{\frac{T}{μ}}$ , where $T$ and $μ$ are the tension and linear density, respectively.

So, the speed of waves on the strings can be calculated by substituting the values given as,

$v_{1} = \sqrt{\frac{T}{μ}} v_{2} = \sqrt{\frac{T}{9 μ}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{9} = 3 \Rightarrow v_{1} = 3 v_{2} . . . (1)$

Also, frequencies of the waves on both the wires would be the same since the wires are joined together. $f_{1} = f_{2} . . . (2)$

(a) When $μ$ is a node, the wave forms on both the wires would be complete segments like in a wire fixed at both ends.

So, $f_{1} = n_{1} \frac{v_{1}}{2 L}$ , where $n_{1} = 1 . 2, 3, 4 . . . . . (3)$

and

$f_{1} = n_{1} \frac{v_{1}}{2 L}$ , where $n_{2} = 1 . 2, 3, 4 . . .$

Substituting in equation $(2)$ , we get

$n_{1} \frac{v_{1}}{2 L} = n_{2} \frac{v_{2}}{2 L}$

Simplifying after substituting the result in equation $(1)$ , we obtain

$\frac{n_{1}}{n_{2}} = \frac{v_{2}}{v_{1}} = \frac{1}{3}$

Hence, it is feasible for all values $n_{1} = 1, 2, 3, . . .$ since feasible values of $n_{2}$ exist for these values.

Hence, upon substituting in $(3)$ and using value of $v_{1}$ , we get the possible frequencies as,

$f_{1} = \frac{1}{2 L} \sqrt{\frac{T}{μ}}, \frac{2}{2 L} \sqrt{\frac{T}{μ}}, \frac{3}{2 L} \sqrt{\frac{T}{μ}} . . .$

(b) when $A$ is an antinode, the frequency of standing waves would be $f = \frac{(2 n - 1) v}{4 L}$

Since, equation $(2)$ would still apply, we can write,

$f_{1} = \frac{(2 n_{1} - 1) v_{1}}{4 L} = \frac{(2 n_{2} - 1) v_{2}}{4 L} = f_{2} . . . (4)$

Substituting equation $(1)$ and simplifying,

$3 (2 n_{1} - 1) = (2 n_{2} - 1) \Rightarrow 6 n_{1} - 2 = 2 n_{2} \Rightarrow n_{2} = 3 n_{1} - 1$

Hence, it is feasible to have $n_{1} = 1, 2, 3, . . .$

Hence, using equation $(4)$ , and values of velocities, the feasible frequencies are,

$\frac{1}{4 L} \sqrt{\frac{T}{μ}}, \frac{3}{4 L} \sqrt{\frac{T}{μ}}, \frac{5}{4 L} \sqrt{\frac{T}{μ}}, . . .$

A light string is tied at one end to fixed support and to a heavy string of equal length $L$ at the other end as shown in figure. Mass per unit length of the strings are $μ$ and $9 μ$ and the tension is $T .$ Find the possible values of frequencies such that point $A$ is a node/antinode.

Important Questions on Superposition of Waves

Learn and Prepare for any exam you want !

A light string is tied at one end to fixed support and to a heavy string of equal length L at the other end as shown in figure. Mass per unit length of the strings are μ and 9μ and the tension is T. Find the possible values of frequencies such that point A is a node/antinode.

Important Questions on Superposition of Waves

Learn and Prepare for any exam you want !

A light string is tied at one end to fixed support and to a heavy string of equal length $L$ at the other end as shown in figure. Mass per unit length of the strings are $μ$ and $9 μ$ and the tension is $T .$ Find the possible values of frequencies such that point $A$ is a node/antinode.