Animated Solution for Physics - Waves: List-I shows four configurations made of straight and semi-circular narrow tubes containing air. A sound wave of wavelength λ=0.29 m enters these structures at the point S and a sound detector is placed at D. Between the points S and D, the sound travels only through the tubes. List-II contains the possible smallest values of l (refer to the figures) for which the detector D records maximum amplitude. Ignore effects of sharp corners. [Given cos(15∘)=0.97]
List-I
(P)
(Q)
(R)
(S)
List-II
(1)
1.32 m
(2)
1.19 m
(3)
0.51 m
(4)
0.29 m
(5)
0.13 m
Select Matching Pairs:
* Multiple Allowed
PMatches
QMatches
RMatches
SMatches
Visualized Solution
\Delta x = n\lambda
For maximum amplitude at the detector, the sound waves from the two paths must interfere constructively.
\Delta x = n\lambda
For the smallest length l, we take n=1.
\Delta x = \lambda = 0.29\text{ m}
\text{Configuration P}
Path 1 (Straight): x1=l
Path 2 (Semi-circle): x2=πr=π(0.5l)=2πl
\Delta x_P = \lambda
\Delta x = x_2 - x_1 = \left(\frac{\pi}{2} - 1\right)l
\left(\frac{\pi}{2} - 1\right)l = 0.29
l = \frac{0.29}{1.57 - 1} = \frac{0.29}{0.57} \approx 0.51\text{ m}
\text{Configuration Q}
Path 1 (Straight): x1=l
Path 2 (Rectangular): x2=0.5l+l+0.5l=2l
\Delta x_Q = \lambda
\Delta x = 2l - l = l
l = \lambda = 0.29\text{ m}
\text{Configuration R}
Path 1 (Straight): x1=l
Path 2 (Vertical + Arc): x2=l+Arc Length
Diameter of semi-circle =l2+l2=l2
Radius R=2l2=2l
\Delta x_R = \lambda
x_2 = l + \pi\left(\frac{l}{\sqrt{2}}\right)
\Delta x = x_2 - x_1 = \frac{\pi l}{\sqrt{2}}
\frac{\pi l}{\sqrt{2}} = 0.29
l = \frac{1.414 \times 0.29}{3.14} \approx 0.13\text{ m}
\text{Configuration S}
Angles of triangle: 45∘,105∘,180∘−(105∘+45∘)=30∘
Let upper path segments be a and b.
By Sine Rule: sin105∘l=sin30∘a=sin45∘b
\Delta x_S = \lambda
Given: sin105∘=cos15∘=0.97
a = l \frac{\sin 30^\circ}{\sin 105^\circ} = l \frac{0.5}{0.97}
b = l \frac{\sin 45^\circ}{\sin 105^\circ} = l \frac{1/\sqrt{2}}{0.97} \approx l \frac{0.707}{0.97}
\text{Final Result}
Path 2 =a+b=l(0.970.5+0.707)≈1.244l
\Delta x = 1.244 l - l = 0.244 l
0.244 l = 0.29 \implies l \approx 1.19\text{ m}
\text{Matching}
P → 3 (0.51 m)
Q → 4 (0.29 m)
R → 5 (0.13 m)
S → 2 (1.19 m)
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The Sigma Insight: Interference of Waves
Solution Diagram
The beauty of physics often lies in its ability to merge seemingly disparate concepts into a single, elegant problem. This question is a perfect example, seamlessly blending the wave nature of sound with classical geometry. Let's embark on a journey through these four distinct configurations to find the perfect length l that creates a symphony of constructive interference.
The Core Principle
Constructive Interference
Before we dive into the shapes, we must understand what the detector at D is looking for. It wants to record a maximum amplitude. This happens when the sound waves traveling through the two different paths arrive perfectly in phase, a phenomenon known as constructive interference.
For this to occur, the difference in the distances they traveled—the path difference Δx—must be an integer multiple of the sound's wavelength λ. Mathematically, Δx=nλ. Since the question asks for the smallest possible value of l, we must look for the first maximum, where n=1. Therefore, our master equation for every configuration is simply:
Δx=λ=0.29 m
Configuration P
The Semi-Circle
Imagine the sound wave splitting at S. One part takes the direct, straight route of length l. The other part takes a scenic detour along a semi-circular tube. The radius of this semi-circle is given as 0.5l.
The length of the straight path is x1=l. The length of the curved path is half the circumference of a circle, x2=πr=π(0.5l)=2πl. The path difference is the difference between these two:
Δx=(2π−1)l
Setting this equal to our wavelength λ=0.29 m:
l=23.14−10.29=0.570.29≈0.51 m
This matches perfectly with entry (3) in List-II.
Configuration Q
The Rectangular Detour
This configuration is straightforward. The upper path forms three sides of a rectangle. The sound travels up by 0.5l, across by l, and down by 0.5l.
The total length of this upper path is x2=0.5l+l+0.5l=2l. The straight path is just x1=l. The path difference is beautifully simple:
Δx=2l−l=l
For the first maximum, l must be exactly equal to the wavelength. Therefore, l=0.29 m. This matches entry (4).
Configuration R
The Tilted Arc
Here is where the geometry gets a bit tricky. The straight path is still l. The upper path, however, goes vertically up by a distance l, and then follows a semi-circular arc to reach D.
Notice the dotted line in the figure. It connects the top of the vertical tube to the detector D. This line is the diameter of the semi-circle. If we look closely, this diameter is the hypotenuse of a right-angled triangle with a base of l and a height of l. By the Pythagorean theorem, the diameter is l2+l2=l2. Consequently, the radius of this semi-circle is R=2l.
The total length of the upper path is the vertical segment plus the arc length:
x2=l+πR=l+2πl
Subtracting the straight path x1=l, the path difference is simply the length of the arc:
Δx=2πl
Equating this to λ and solving for l:
l=π2×0.29=3.141.414×0.29≈0.13 m
This corresponds to entry (5).
Configuration S
The Triangular Path
Our final configuration forms a triangle. The straight path is the base, with length l. The upper path consists of the other two sides of the triangle. We are given two angles: 45∘ and 105∘. Since the sum of angles in a triangle is 180∘, the third angle must be 30∘.
To find the lengths of the upper segments (let's call them a and b), we deploy the Law of Sines:
sin105∘l=sin30∘a=sin45∘b
The problem graciously provides cos15∘=0.97. Thanks to the trigonometric identity sin(90∘+θ)=cosθ, we know that sin105∘=cos15∘=0.97. Now we can solve for a and b:
a=lsin105∘sin30∘=l0.970.5
b=lsin105∘sin45∘=l0.971/2≈l0.970.707
The total length of the upper path is a+b≈l(0.971.207)≈1.244l. The path difference is:
Δx=1.244l−l=0.244l
Setting this equal to λ=0.29 m:
l=0.2440.29≈1.19 m
This matches entry (2).
The Final Match
Bringing it all together, we have successfully mapped every configuration:
P matches with 3Q matches with 4R matches with 5S matches with 2
This is a masterclass in applying fundamental physics principles to diverse geometric setups. Always remember to break down complex paths into simple, calculable segments!
