Question

A perfect elastic string of length l which is under tension T and its fixed at both ends, has a linear mass density. The string is given initial deflection and initial velocity at its various points and is released at time t=0.The string execute small transverse vibrations. The initial deflection and initial velocity of the spring at any point x are denoted by h(x) and v(x) respectively. Find the different normal modes of vibrations and deflection of the string at any point x at t=0

Answer 1

Answer:

lcx gvigj gn jxgzrzb g

Explanation:

xxyxugyfucbh b hb jb n

Answer 2

Given:

String length = l

Fixed at both ends.

Tension = T

Linear mass density = u

Mass of length = ul

To Find:

The different normal modes of vibrations and deflection of the string at any point x at t=0

Solution:

Given the ends of the string are fixed.

Then,

• At the ends, the displacement of the string should be zero.
• A transverse wave propagating along the string towards the other fixed end will be reflected back.

• These two waves add up to produce the displacement in the string.

• These two waves travelling in opposite directions bounce back and forth between the ends.
• Y( x, t ) = ym sin(kx -ωt) (Forward)+ ym sin(kx +ωt)(Backward)
• Y(x,t) = 2ym sin(kx) cos(ωt)

A string fixed at one end will produce nodes at :

• We know if ∧ = wavelength,
• For first mode ,
• l  = ∧/2
• f∧ = V , V = Speed of wave
• f = V/2l
• For nth mode,
• f = nV/2l

Therefore prorogation constant k = ω/V = 2πf/V = nπ/l

Therefore, wave equation for deflection at any point x :

• Y(x,t ) = ym sin (nπx/l ) cos (2πft )

At t = 0 ,

• Y(x , 0 ) = ym sin ( nπx/l)

The different normal modes of vibrations are at x = l/n and deflection of the string at any point x at t=0 Y(x , 0 ) = ym sin ( nπx/l).