Question

. Derive an expression for the natural frequency of the free longitudinal vibration by (i)

Equilibrium method (ii) Energy method (iii) Rayleigh’s method​

Answer 1

An expression for the natural frequency of the free longitudinal vibration :

Explanation:

 The natural Frequency of the Free Longitudinal Vibrations is determined by the following methods:

1. Equilibrium Method
2. Energy Method
3. Rayleigh’s method

i) Equilibrium Method :

• Let's consider a spring with a mass in an unstrained position as shown in the below fig.
• The consider that the mass of the spring is negligible.

• Suppose

s = Stiffness of the constraint

m = Mass of the body  

W = Weight of the body  

δ = Static deflection of the spring  

• Now, the body is in the equilibrium position, the gravitational pull W = m.g is balanced by a force of spring, such that W = s. δ

• Now, the displacement to the mass m by a distance x from its equilibrium position.

• The restoring force will be

          = $W - s( \delta + x ) = W - s\delta - sx = -s x$                        ( ∵ W = s. δ)

• Taking upward force as negative

• Accelerating force = Mass × Acceleration

$\displaystyle m \times \frac{d^2x}{dt^2}$

• Take downward force as positive.

• From the above two equations of motion of the body of mass m after time t is calculated by

$\displaystyle m \times \frac{d^2x}{dt^2} = -s \times x$

$\displaystyle m \times \frac{d^2x}{dt^2} +s \times x =0$

$\displaystyle \frac{d^2x}{dt^2} +\frac{-s \times x}{m} =0$

• From the above fundamental equation of the simple harmonic motion of the body is

$\displaystyle \frac{d^2x}{dt^2}+\omega^2x=0$

• So equating above these two similar equations, will get,

$\displaystyle \omega = \sqrt{\frac{s}{m} }$

• The time period is given by

$\displaystyle t_p = \frac{2\pi }{\omega}$

• The natural frequency is given by

$\displaystyle\mathbf{ f = \frac{1}{t}=\frac{1}{2\pi } \sqrt{\frac{s}{m} }}$

 

ii) Energy Method :

• In the free vibrations, no energy is transferred to the system or from the system, so that the total Kinetic energy and the potential energy of the system must be a constant quantity which is the same at all times.

$\displaystyle \frac{d}{dt}(K.E. +P.E)=0$

• We will find kinetic energy and potential energy respectively,

$\displaystyle K.E.= \frac{1}{2}\ mv^2 = \frac{1}{2}m(\frac{dx}{dt})^2$

$\displaystyle P.E.= (\frac{0+sX}{2})X=\frac{1}{2}s\times X^2$

• Now, put these two equations in the above equation, by solving them we get

$\displaystyle \frac{d^2x}{dt^2}+\frac{s}{m}\times X =0$

• The fundamental equation of the simple harmonic motion of the body is calculated by

$\displaystyle \frac{d^2x}{dt^2} +\omega^2 x = 0$

• So equating these two similar equations will get  

$\displaystyle \omega = \sqrt{\frac{s}{m} }$

• The natural time period is given by

$\displaystyle t_p = \frac{2\pi }{\omega}$

• The natural frequency is given by

$\displaystyle\mathbf{ f = \frac{1}{t}=\frac{1}{2\pi } \sqrt{\frac{s}{m} }}$

iii) Rayleigh’s Method :

• The maximum kinetic energy at the mean position is equal to the hightest potential energy or also known as the strain energy at the extreme position.

• Suppose the motion executed by the vibration to be simple harmonic then,

x = X sin ω.t

Where

x = The displacement of the body from the mean position next to time  

X = Maximum displacement of the body from mean position to an extreme position

• Differentiating the equation x = X sin ω.ton both sides,we get,

$\displaystyle \frac{dx}{dt} = \omega \times x \ cos\omega\ t$

• At the mean position, t = 0, then maximum velocity at the mean position is

$\displaystyle v = \frac{dx}{dt}=\omega\ \times\ X$

• The maximum Kinetic energy at the mean position is given by

$\displaystyle K.E. = \frac{1}{2}mv^2=\frac{1}{2}(\frac{dx}{dt})^2 = \frac{1}{2}\ m\ \omega^2\ X^2$

• The maximum potential energy at the extreme position is given by

$\displaystyle P.E. =( \frac{0 +s.X}{2}) X=\frac{1}{2}\ s\ X^2$

• Equating the above two equations, then solving the above equation we get

$\displaystyle \omega = \sqrt{\frac{s}{m} }$

• Time period is

$\displaystyle t_p = \frac{2\pi }{\omega}$

• The natural frequency is calculated by

$\displaystyle\mathbf{ f = \frac{1}{t}=\frac{1}{2\pi } \sqrt{\frac{s}{m} }}$

Refer to the following attachment: