# 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

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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}

^{-6}m. Write the equation of at a point 2 cm from the mid – point and those of the constituent waves in the rod.

^{11}Nm

^{-2}; density = 8000kgm

^{-3}

## 3 Answers

^{11}/8 x 10

^{3}= 5000 m/s

^{-6}m

^{-6}m ⇒ a = 10

^{-6}m

_{1}= a sin 2π/ℷ (vt – x)

_{1}= 10

^{-6}sin 2π/0.4 (5000t – x)

_{2}= 10

^{-6}sin[ 2π/0.4 (5000t + x]

^{-6}cos (2π/0.4 x) sin (2π/0.4 x 5000t)

^{-6}cos (2π/0.4 x 0.02 ) sin (2π/0.4 x 5000t)

^{-6}cos (0.1 π) sin (25000 π t)

Thanks

Aditi Chauhan

askIITians Faculty

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