Question-
A slab whose dielectric constant is K, is attached to a string of linear mass density u, while its one end is fixed to a rigid support. Its cross section has sqaure plates of side b and seperation between them is d. Find the fundamental frequency of the vibration of the string.
d
Aϵ
0
C
′
=
d−t+
k
t
Aϵ
0
Put t=
4
3d
;
C
′
=
3+k
4k
.
d
Aϵ
0
C
′
=
3+k
4k
.C
Answers
The equation of a travelling wave propagating along the positivey-direction is given by the displacement equation: ...(i) Linear mass density, Frequency of the tuning fork, Amplitude of the wave, ...(ii) Mass of the pan, m Tension in the string, The velocity of the transverse wave v, is given by the relation: Angular Frequency, ...(iii) Wavelength, Propagation constant, ...(iv) Substituting the values from equations (ii), (iii), and (iv) in equation (i), we get the displacement equation: .
Let us write the given data and solve the question
Density = U
Cross section of plates = b
Separation = d
Fundamental frequency (n) = ?
SOLUTION:
Let us use that formula here
n = 1/2L √T/μ
As we already know the value of T from the given data
Let us try to derive the value of T
T = -dU/dx = 1/2(k-1)E₀ bV²/ d equation 1
Let us apply the value of T in frequency formula (n)
Then,
n = 1/2L √(k-1)E₀bV² /2dμ
This is final equation