Question

A taut string of density p and length b is fixed at both ends. At a distance 3b/7 from the origin it is pulled up by an amount h, and at a distance 4b/7 it is pulled down by the same amount. Both points are released simultaneously.

Describe the motion and the amplitudes of the normal modes (So, which modes are not excited and why not)

Answer 1

When the string is moved up wave pulse on the string move fron left to right ie in +ve x- direction represented by:

y₁ = r sin 2*pie/ lambda( vt +x )                        ----------------1

here r = amplitude, lambda = wavelength, v= velocity, t = time

When the string is pulled down the wave pulse will move in opposite direction as above and given by

y_{2 =} r sin [ 2*pie/ lambda( v t - x) + pie)                 ( since between two pulses phase difference is pie)

y₂ = - r sin [2*pie/ lambda (vt - x) ]                       --------------2

according to superposition principle resultant displacement is given by:

y = y₁ + y₂

y = r [ sin {2*pie/ lambda( vt+x) } - sin {2*pie/ lambda (v t - x) ]

noe by using formula sin C - sin D = 2 cos( C+D/2 )*sin ( C-D/2)

y = 2r cos [(2*pie/ lambda)*v t] * sin ( 2*pie/ lambda) *x       ------------3 ( this is the resultant amplitude)

At one end of string one x= 0 , then from eq. 3

y = 0

at x = b, y = 0 since other end of string is also fixed

sin 2*pie/ lambda* b= 0 = sin n*pie                           ( sin n*pie = 0)

lambda = 2 b/ n

here n = 1,2,3 -------- are normal modes of vibration of string.

For first normal mode , lambda 1 = 2 b

b = lambda/ 2

And frequency of vibration in this mode is given by,

f = velocity/ wavelength( lambda 1)

f = v / 2b

similarly we can find for other modes.