Animated Solution for Physics - Waves: Two uniform strings of mass per unit length μ and 4μ, and length L and 2L, respectively, are joined at point O, and tied at two fixed ends P and Q, as shown in the figure. The strings are under a uniform tension T. If we define the frequency ν0=2L1μT, which of the following statement(s) is(are) correct?
Select Answer:
* Multiple Correct
Visualized Solution
v=μT
v1=μT
v2=4μT=21μT
v2=2v1
f1=f2=f
f=λ1v1=λ2v2
λ2λ1=v2v1=2
λ1=2λ2
Case 1: Node at O
String 1: L=m2λ1
String 2: 2L=n2λ2
Solving for m and n
2LL=nλ2/2m(2λ2)/2
21=n2m⟹n=4m
For fmin, choose minimum integers:
m=1,n=4
\text{Minimum Frequency & Nodes}
λ1=12L=2L
fmin=λ1v1=2L1μT=ν0
Total Nodes=(m+1)+(n+1)−1
Total Nodes=1+4+1=6
Case 2: Antinode at O
String 1: L=(2m−1)4λ1
String 2: 2L=(2n−1)4λ2
The Mathematical Contradiction
2LL=(2n−1)λ2/4(2m−1)(2λ2)/4
21=2n−12(2m−1)
2n−1=8m−4
2(n−4m)=−3
Conclusion
Even=Odd
No integer solutions exist.
No such mode is possible.
The Way Forward
Master the boundary conditions!
What if μ2=9μ?
What if O was attached to a spring?
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The Sigma Insight: Standing Waves in Strings and Organ Pipes
Solution Diagram
The Anatomy of a Composite String
Imagine a guitar string, but instead of being uniform, it's spliced together from two completely different materials. This is the essence of our problem. We have a lighter string of length L and mass density μ joined to a heavier string of length 2L and mass density 4μ.
Because they are pulled taut by the exact same tension T, the wave speed will drastically differ between the two sections. The wave speed in a string is governed by the equation v=μT. For the lighter string, the speed is v1=μT. For the heavier string, the speed drops to v2=4μT=2v1. The wave travels twice as fast in the lighter section!
The Universal Law of Frequencies
For a standing wave to successfully form across this entire composite system, the two strings must "agree" on a single frequency of vibration. If they vibrated at different frequencies, the junction O would tear itself apart.
Since frequency f is constant, and f=λv, the wavelengths must adjust to compensate for the different speeds. We find that λ2λ1=v2v1=2. This tells us that the wavelength in the lighter string is exactly twice the wavelength in the heavier string.
Case 1
The Node at the Junction
Let's test the first boundary condition: what if the junction O is a node? Since the far ends P and Q are rigidly fixed, they are also nodes. This means both strings must fit an integer number of half-wavelengths perfectly within their lengths.
For the first string: L=m2λ1.
For the second string: 2L=n2λ2.
Substituting our golden rule λ1=2λ2 into these equations and dividing them yields a beautiful, clean ratio: 21=n2m, which simplifies to n=4m.
To find the minimum frequency, we need the longest possible wavelengths, which corresponds to the smallest possible integers for m and n. Thus, we choose m=1 and n=4.
Counting the Invisible
Nodes and Loops
With m=1, the fundamental wavelength of the first string is λ1=2L. Plugging this back into our frequency equation gives fmin=2Lv1=2L1μT, which is exactly $
u_0$. Option (A) is correct.
Now, let's count the nodes. The left string has 1 loop, and the right string has 4 loops. A string with k loops has k+1 nodes. However, because they share the node at junction O, we must subtract 1 to avoid double-counting. The total number of nodes is (1+1)+(4+1)−1=6. Option (C) is also correct.
Case 2
The Antinode Paradox
What if the junction O was an antinode (a point of maximum displacement)? In this scenario, each string has a node at one end and an antinode at the other. Their lengths must now be odd multiples of their quarter-wavelengths.
For the first string: L=(2m−1)4λ1.
For the second string: 2L=(2n−1)4λ2.
The Mathematical Impossibility
Let's perform the exact same substitution and division as before. We substitute λ1=2λ2 and divide the equations:
2LL=(2n−1)λ2/4(2m−1)(2λ2)/4
This simplifies to 21=2n−12(2m−1). Cross-multiplying gives 2n−1=8m−4, which rearranges to 2(n−4m)=−3.
Look closely at this final equation. The left side, 2(n−4m), is inherently an even number. The right side, −3, is an odd number. An even number can never equal an odd number!
This mathematical contradiction is profound. It proves that the physical boundary conditions clash entirely. It is physically impossible for this specific composite string to sustain a standing wave with an antinode at the junction. Therefore, Option (D) is correct.
