Wave Motion> A metallic rod of length lm is rigidly cl...

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

Simran Bhatia , 10 Years ago

Grade 11

3 Answers

Aditi Chauhan

Last Activity: 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

∴ 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

Last Activity: 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

Last Activity: 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