Question

Explain the formation of stationary waves in stretched strings and hence deduce the laws of transverse waves in stretched strings TS J-16,18 AP J-18

Answer 1

Hint: To solve this type of question we will first discuss how the stationary waves will be formed and them we will deduce the law of transverse waves in stretched strings by using analytical method and then taking four cases so as to know about nodes and antinodes.

Complete step-by-step solution:
Let us first discuss how stationary waves are.
So, stationary waves are formed when two progressive waves of same amplitude and same wavelength that are travelling along a straight line in opposite directions when they are superimposed on each other.
Now, we will deduce the law of transverse waves in stretched strings by analytical method:
Let there be a progressive wave of amplitude a and wavelength λ which is travelling in the direction of X axis.
\[{{\text{y}}_{1}}=\text{ a Sin 2}\pi [\text{ }\dfrac{t}{T}\text{- }\dfrac{x}{\lambda }]\text{ }\]Let this be equation (1)
If this wave given is reflected from a free end and it travels in the opposite that is negative direction of X axis, then
\[{{\text{y}}_{2}}=\text{ a sin 2}\pi [\dfrac{t}{T}\text{ + }\dfrac{x}{\lambda }]\text{ }\]Let this be equation (2)
Both the waves will superimpose each other so the resultant will be:
\[y={{y}_{1}}+{{y}_{2}}\]
\[=\text{a }\!\![\!\!\text{ sin 2}\pi \text{ (}\dfrac{t}{T}-\dfrac{x}{\lambda })+sin2\pi \text{(}\dfrac{t}{T}+\dfrac{x}{\lambda })]\]
\[=\text{a }\!\![\!\!\text{ 2sin (}\dfrac{2\pi t}{T})\text{ cos (}\dfrac{2\pi x}{\lambda })]\]
So, here we can say that,
 \[=2a\text{sin (}\dfrac{2\pi t}{T})\text{ cos (}\dfrac{2\pi x}{\lambda })\] Let this equation be (3)
The above equation is the equation of stationary waves.
Let us take the first case,
When\[x=0,\dfrac{\lambda }{2},\lambda ,\dfrac{3\lambda }{2}\], then value of \[\cos \dfrac{2\pi x}{\lambda }=\pm 1\]
So, at this point we have a resultant maximum amplitude which is called Antinodes.
For second case,
When\[x=\dfrac{\lambda }{4},\dfrac{3\lambda }{4},\dfrac{5\lambda }{4}\], then value of \[Cos\dfrac{2\pi x}{\lambda }=0\]
Here the resultant amplitude is 0, so this position is called Nodes.
The distance between any two successive antinodes and nodes is equal to \[\dfrac{\lambda }{2}\]and the distance between any antinode and node is \[\dfrac{\lambda }{4}\].
For Third case,
Where t = \[0,\dfrac{T}{2},T,\dfrac{3T}{2}\]then \[\sin \dfrac{2\pi t}{T}=0\]
So, the displacement here will be 0.
For fourth case,
When\[t=\dfrac{T}{4},\dfrac{3T}{4},\dfrac{5T}{4}\], then\[\sin \dfrac{2\pi t}{T}=\pm 1\]
So, the displacement here will be maximum.

Note: In the question given above always remember that the resultant of maximum amplitude is antinodes and the resultant of minimum amplitude is nodes. And also, the distance between two successive antinodes and nodes is always equal to \[\dfrac{\lambda }{2}\]and the distance between any node and antinodes is\[\dfrac{\lambda }{4}\].