Question

A block is placed on a frictionless horizontal table. The mass of the block is m and springs are attached on either side with force constants\[{k_1}\]and\[{k_2}\]. If the block is displaced a little and left to oscillate, then the angular frequency of oscillation will be
A. \[{\left(\dfrac {{k_1}+{k_2}}{m}\right)}^{\frac {1}{2}}\]
B. \[{\left(\dfrac {{k_1}{k_2}}{m\left({k_1}+{k_2}\right)}\right)}^{\frac {1}{2}}\]
C. \[{\left(\dfrac {{k_1}{k_2}}{m\left({k_1}-{k_2}\right)}\right)}^{\frac {1}{2}}\]
D. \[{\left(\dfrac {{k_1}^2+{k_2}^2}{m\left({k_1}+{k_2}\right)}\right)}^{\frac {1}{2}}\]

Answer 1

Hint:We find the equivalent spring constants for the combination of the springs attached to the block and then use the formula of the angular frequency to find the angular frequency of the harmonic motion of block-spring system.

Formula used:
\[\dfrac{1}{{{k_{eq}}}} = \dfrac{1}{{{k_1}}} + \dfrac{1}{{{k_2}}} + \dfrac{1}{{{k_3}}} \ldots + \dfrac{1}{{{k_n}}}\]
Where \[{k_{eq}}\]is the equivalent spring constant when the springs are connected in series.

\[{k_{eq}} = {k_1} + {k_2} + {k_3} \ldots + {k_n}\],
Where \[{k_{eq}}\]is the equivalent spring constant when the springs are connected in parallel.

\[\omega = \sqrt {\dfrac{{{K_{eq}}}}{m}} \],
Where \[\omega \]is the angular frequency of spring-mass system.

Complete step by step solution:

Fig: The springs attached to the mass m
As the springs connected on either side of the mass, so when the mass is displaced on either side the change in length of the springs will be same. Hence, the springs are connected in parallel.

The equivalent spring constant of the given combination will be,
\[{k_{eq}} = {k_1} + {k_2}\]
The mass of the block is given as m.

Using the angular frequency formula for the spring block system, we get
\[\omega = \sqrt {\dfrac{{\left( {{k_1} + {k_2}} \right)}}{m}} \]
\[\omega = {\left(\dfrac {{k_1}+{k_2}}{m}\right)}^{\frac {1}{2}}\]

So, the angular frequency of the given spring-block system is \[{\left(\dfrac {{k_1}+{k_2}}{m}\right)}^{\frac {1}{2}}\]

Therefore, the correct option is (A).

Note: We should be careful while analysing the combination of the springs in order to find the equivalent spring constant. If the change in length of the springs was different when the mass is displaced a little then the combination will be in series.