
Two identical springs of constant K are connected in series and parallel as shown in figure. A mass m is suspended from them. The ratio of their frequencies of vertical oscillations will be

A. 2:1
B. 1:1
C. 1:2
D. 4:1
Answer
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Hint: If two springs of constant \[{k_1}\] and \[{k_2}\] respectively is connected in parallel combination then the effective constant of the system will be\[{k_p} = {k_1} + {k_2}\] and for a series combination of the springs its will be \[{k_s} = \dfrac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}\].
Formula used:
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \]
Where, f = Frequency of oscillation, k= Springs constant and m = Mass suspended to spring
Complete step by step solution:
Given here are two spring-mass systems each having a combination of two springs of constant k, we have to find the ratio of their frequencies.
The frequency of oscillation is given by,
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \,.......(1)\]
Where, f = Frequency of oscillation, k= Springs constant and m = Mass suspended to spring
In figure A two springs each having constant k are connected in series combination and the effective constant of the system will be,
\[{k_s} = \dfrac{{kk}}{{k + k}} \Rightarrow {k_s} = \dfrac{k}{2}\]

Image 1 : Equivalent diagram of system A
Using equation (1) frequency of oscillation of the system in figure A will be,
\[{f_A} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{{2m}}} \,.......(2)\]
The effective constant for two springs of constant k each in figure B will be,
\[{k_p} = k + k = 2k\]

Image 2 : Equivalent diagram of system B
The frequency of oscillation of the system will be,
\[{f_B} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{2k}}{m}} \,.......(3)\]
From equations (2) and (3) we get,
\[\dfrac{{{f_A}}}{{{f_B}}} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{{2m}}} \, \times 2\pi \sqrt {\dfrac{m}{{2k}}} \,\]
The above equation can be written as,
\[\dfrac{{{f_A}}}{{{f_B}}} = \sqrt {\dfrac{k}{{2m}} \times \dfrac{m}{{2k}}} \Rightarrow \dfrac{{{f_A}}}{{{f_B}}} = \dfrac{1}{2}\]
Hence, the ratio of frequency is 1:2. Thus, option C is the correct option.
Note: If the external force is acting on a series combination of springs then the force is the same on each spring without any change in magnitude and the deformation of the system will be the sum of individual deformation of each spring. In parallel combinations of the springs, deformation of the system is common deformation among all springs.
Formula used:
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \]
Where, f = Frequency of oscillation, k= Springs constant and m = Mass suspended to spring
Complete step by step solution:
Given here are two spring-mass systems each having a combination of two springs of constant k, we have to find the ratio of their frequencies.
The frequency of oscillation is given by,
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \,.......(1)\]
Where, f = Frequency of oscillation, k= Springs constant and m = Mass suspended to spring
In figure A two springs each having constant k are connected in series combination and the effective constant of the system will be,
\[{k_s} = \dfrac{{kk}}{{k + k}} \Rightarrow {k_s} = \dfrac{k}{2}\]

Image 1 : Equivalent diagram of system A
Using equation (1) frequency of oscillation of the system in figure A will be,
\[{f_A} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{{2m}}} \,.......(2)\]
The effective constant for two springs of constant k each in figure B will be,
\[{k_p} = k + k = 2k\]

Image 2 : Equivalent diagram of system B
The frequency of oscillation of the system will be,
\[{f_B} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{2k}}{m}} \,.......(3)\]
From equations (2) and (3) we get,
\[\dfrac{{{f_A}}}{{{f_B}}} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{{2m}}} \, \times 2\pi \sqrt {\dfrac{m}{{2k}}} \,\]
The above equation can be written as,
\[\dfrac{{{f_A}}}{{{f_B}}} = \sqrt {\dfrac{k}{{2m}} \times \dfrac{m}{{2k}}} \Rightarrow \dfrac{{{f_A}}}{{{f_B}}} = \dfrac{1}{2}\]
Hence, the ratio of frequency is 1:2. Thus, option C is the correct option.
Note: If the external force is acting on a series combination of springs then the force is the same on each spring without any change in magnitude and the deformation of the system will be the sum of individual deformation of each spring. In parallel combinations of the springs, deformation of the system is common deformation among all springs.
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