Question

A mass $M$ supported by a spring has a static deflection of $\delta $. The frequency of oscillation is
A. $\dfrac{1}{{2\pi }}\sqrt {\dfrac{\delta }{M}} $
B. $\dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{\delta }} $
C. $\dfrac{1}{{2\pi }}\sqrt {g\delta } $
D. $\dfrac{1}{{2\pi }}\sqrt {\dfrac{{M\delta }}{g}} $

Answer 1

Hint:First of all, we have to find the normal frequency of a spring-mass system under free vibration without damping. Then we have to substitute the static deflection and find the frequency of oscillation in this case.

Complete step by step answer:
The static deflection of a spring-mass system is defined as the deflection of spring constant $k$ as a result of the gravity force of the mass or weight.
So static deflection can be represented as,
$\delta = \dfrac{{Mg}}{k} - - - - \left( 1 \right)$ where $\delta $ is the static deflection, $Mg$ is the weight of the body or the gravity force on mass and $k$ is the spring constant.
From Newton’s equation let us consider the expression of free vibration,
$Mg = - kx - - - - \left( 2 \right)$ where $ - ve$ sign represents that the spring force acts opposite to the weight exerted by the body.
Angular frequency ${\omega _n}$ of a spring mass system is expressed as,
${\omega _n} = \sqrt {\dfrac{k}{M}} - - - - \left( 3 \right)$
Again, we know that time period to complete one sinusoidal oscillation is $T = \dfrac{{2\pi }}{{{\omega _n}}} - - - - - \left( 4 \right)$

Now, the natural frequency $f$ of the body is represented as,
$f = \dfrac{1}{T} - - - - - \left( 5 \right)$
Substituting the value of $T$ from equation $\left( 4 \right)$ we get,
$f = \dfrac{1}{T} = \dfrac{1}{{\dfrac{{2\pi }}{{{\omega _n}}}}} - - - - - \left( 6 \right)$
Again, substituting the value of ${\omega _n}$ from equation $\left( 3 \right)$ we have,
$f = \dfrac{1}{{\dfrac{{2\pi }}{{{\omega _n}}}}} = \dfrac{{{\omega _n}}}{{2\pi }} = \dfrac{{\sqrt {\dfrac{k}{M}} }}{{2\pi }}$
So, the equation can be represented as,
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{M}} - - - \left( 7 \right)$
Multiplying $\sqrt g $ to both the upper and lower part of the right side of equation $\left( 7 \right)$ we get,
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{kg}}{{Mg}}} - - - - \left( 8 \right)$
Putting the value of $\delta $ from equation$\left( 1 \right)$ to equation $\left( 8 \right)$ we get,
$\therefore f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{\delta }} $
So, the frequency of oscillation is $f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{g}{\delta }}$.

Hence, the correct option is B.

Note: It must be noted that we have to consider the natural free vibration of a body as a generalised equation to find the angular natural frequency. Forced vibration of damped vibration must not be considered in this case. Static deflection cannot be used in case of rubber or other elastic material as they exhibit a dynamic stiffness which differs from static stiffness.