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A metallic rod of length lm is rigidly clamped at its mid point. Longitudinal stationary waves are set up in the rod in such a way that there are two nodes on either side of the mid – point. The amplitude of an antinode is 2 x 10 -6 m. Write the equation of at a point 2 cm from the mid – point and those of the constituent waves in the rod. (Young’s Modulus of the material of the rod = 2 x 10 11 Nm -2 ; density = 8000kgm -3

A metallic rod of length lm is rigidly clamped at its mid point. Longitudinal stationary waves are set up in the rod in such a way that there are two nodes on either side of the mid – point. The amplitude of an antinode is 2 x 10-6 m. Write the equation of at a point 2 cm from the mid – point and those of the constituent waves in the rod.
 (Young’s Modulus of the material of the rod = 2 x 1011 Nm -2; density = 8000kgm-3

Grade:11

3 Answers

Aditi Chauhan
askIITians Faculty 396 Points
10 years ago
Hello Student,
Please find the answer to your question
The placements of the nodes and antinodes on the rod are shown in the figure
236-2300_6.png
∴ ℷ + ℷ/4 = 0.5 ⇒ ℷ = 0.4m
Also, the velocity of waves produced in the rod,
v = √Y/p = √2 x 1011/8 x 103 = 5000 m/s
Since, amplitude of antinodes = 2 x 10-6 m
∴ 2a = 2 x 10-6 m ⇒ a = 10-6 m
The equation of wave moving in the positive X – direction will be
y1 = a sin 2π/ℷ (vt – x)
⇒ y1 = 10-6 sin 2π/0.4 (5000t – x)
The equation of wave after reflection and moving in X – axis is
y2 = 10-6 sin[ 2π/0.4 (5000t + x]
The equation of the stationary wave is
y = 2a cos 2π/ℷ x sin 2π/ℷ vt
∴ y = 2 x 10-6 cos (2π/0.4 x) sin (2π/0.4 x 5000t)
Equation of wave at x = 2 cm
y = 2 x 10-6 cos (2π/0.4 x 0.02 ) sin (2π/0.4 x 5000t)
y = 2 x 10-6 cos (0.1 π) sin (25000 π t)

Thanks
Aditi Chauhan
askIITians Faculty
sumi
11 Points
6 years ago
Hi Aditi.
In your answer 
The equation of the stationary wave is
y = 2a cos 2π/ℷ x sin 2π/ℷ vt
but here if i take x=0 (which is mid point, by going your logic) then Amplitude of particle become 2a while in question it is node so amplitude should be 0
Actually if we take x=0 at midpoint (Node) then equation of stationary wave should be
The equation of the stationary wave is
y = 2a sin 2π/ℷ x cos 2π/ℷ vt
Rishi Sharma
askIITians Faculty 646 Points
4 years ago
As found in case of string , in case of rods also clamped point behaves as a

node while the free end antinode . the situation is shown in figure i have

made for you in paint : -)

Because the distance between two consecutive nodes is ( lambda/2)

while between a node and antinode is lambda/4hence

4× { lamda/2} + 2 {lambda/4} = > L or lambda = > 2×1/5 = > 0.4m

further , it is given that

lambda = > 2×10^11 N/m² andρ = > 8×10³kg/m³

v= >√(λ/ρ) = >√[(2×10^11)/8×10^3] = > 5000 m /s

Hence fromν = > nλ

n = > v/λ = > 5000/0.4 = > 125 Hz

Now if incident and reflected waves along the rod are

y1 = > A sin(ωt -kx) and y2 = >sin(ωt+kx+Φ)

THE RESULTANT WAVE WILL BE

y = > y1 +y2 => Asin (ωt-kx) + sin(ωt+kx+Φ) = > 2A cos ( kx + (Φ/2) )sin (ωt+(Φ/2)
Dear Student,
Please find below the solution to your problem.

Because there is an antinode at the free end of the rod , hence amplitude is maximum at x = > 0 so
cos ( k×0 + (Φ/2) ) = > maximum = > 1 i.e Φ = > 0
And ,
A max = > 2A = > 2×10^-6 m ------------------------( given )
y = > 2×10^-6 cos kx sinωt
as we know that y => 2×10^-6 cos 5πx sin 25000πt
Now because for a point 2cm from the midpoint x = > (0.50+,-0.02) ,
and hence we got
y => 2×10^-6 Cos 5π(0.5 +,- 0.02) sin25000(πt) is our required answer

Thanks and Regards

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