Question

A metallic rod of length 1 m us rigidly clamped at its midpoint . Longitudinal stationary waves are set up in the rod in such a way that there are two nodes on either side of the midpoint . the amplitude of an antinode in 2*10^-6 m . write the equation of motion at a point 2 cm from the midpoint and those of constituent waves in the rod.

{ y= 2*10^11 N/m^2 and rough = > 8*10^3 kg/m^3

Answer 1

Hello pari : - ) 

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/4 hence 

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  and ρ = > 8  ×10 ³kg/m³

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

Hence from ν = > 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)  

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 

HOPE THIS HELPS YOU ! 

@ engineer gopal khandelwal