Question

A mass m is suspended from a spring of length l and force constant K. The frequency of vibration of the mass is ${f_1}$. The spring is cut into two equal part and the same mass is suspended from one of the parts. The new frequency of vibration of mass is ${f_2}$. Which of the following relations between the frequencies is correct
A. ${f_1} = \sqrt 2 {f_2}$
B. ${f_1} = {f_2}$
C. ${b^2} < 4ac$
D. ${f_2} = \sqrt 2 {f_1}$

Answer 1

Hint:This problem is based on the concept of oscillations and waves section of physics. First we have to find the spring constant after the reduced mass. Then taking the ratio of the frequencies will give you the answer. The ratios of frequencies are calculated using the frequency equation.

Formula Used:
The equation of frequency:
 $f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} $
Where $f$= frequency, $k$= force constant and $m$= mass.

Complete step by step solution:
In the first case the values are,
$m$ = mass, ${f_1}$=frequency, $l$ = length, ${k_1}$= force constant (${k_1} = k$).
In the second case with reduced length the values are,
$m$ = mass, ${f_1}$=frequency, $\dfrac{l}{2}$ = length, ${k_2}$= force constant.
We need to find the value of force constant (${k_2}$). We know the relation between length of a spring and its force constant.
$k \propto \dfrac{1}{l}$
Here, the length of the spring is $\dfrac{l}{2}$. Then the force constant is ${k_2} = 2k$.

The length of the spring is decreased by the factor of 2 and then the force constant will increase by the factor of 2. The equation of frequency is given by
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} $
Where $f$= frequency, $k$= force constant and $m$= mass.

The frequency for the first case,
\[{f_1} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{{k_1}}}{m}} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \\ \]
The frequency for the first case,
\[{f_2} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{{k_2}}}{m}} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{2k}}{m}} \\ \]
Taking the ratio between \[{f_1}\] and \[{f_2}\]:
\[\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} }}{{\dfrac{1}{{2\pi }}\sqrt {\dfrac{{2k}}{m}} }} \\
\Rightarrow \dfrac{{{f_1}}}{{{f_2}}} = \sqrt {\dfrac{k}{m}} \times \sqrt {\dfrac{m}{{2k}}} \]
\[\Rightarrow \dfrac{{{f_1}}}{{{f_2}}} = \dfrac{1}{{\sqrt 2 }}\]
\[\therefore {f_2} = \sqrt 2 {f_1}\]

Hence, the correct option is option D.

Note: The diameter of the coil turns, the number of turns per unit length, the total length of the spring, and the stiffness of the spring material will all affect the spring constant. It also depends on the thickness of the wire from which the spring is wound.