Question

A 1 Kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in a parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8kg block is;
A. $ \dfrac{1}{4} $ Hz
B. $ \dfrac{1}{2\sqrt{2}} $ Hz
C. 2Hz
D. $ \dfrac{1}{2} $ Hz

Answer 1

Hint: Two springs are oriented parallel to each other. The equivalent spring constant of the system is the summation of spring constants of the individual spring. Time period of one oscillation is calculated for both the systems.
Formula used:
 $ {{T}_{1}}=2x\sqrt{\dfrac{{{m}_{1}}}{k}}=\dfrac{1}{f} $
 $ {{k}_{eq}}=k+k=2k $

Complete step-by-step answer:
In engineering and physics, a spring system or spring network is a model of physics described as a graph with a position at each vertex and a spring of given stiffness and length along each edge. This generalized Hooke's law to higher dimensions.
.Frequency of the 1kg block attached to a spring is given to us as 1Hz.
So $ {f_1} $ =1Hz
Time period of oscillation of the single spring system is
 $ {{T}_{1}}=2x\sqrt{\dfrac{{{m}_{1}}}{k}}=\dfrac{1}{f} $
On substituting the values in the above equation we get,
 $ 2x\sqrt{\dfrac{1}{k}}=1 $
 $ \Rightarrow k=4{{x}^{2}} $ ……… (1)
In the parallel spring system,
m2=8kg
 $ {{k}_{eq}}=k+k=2k $
 $ T=2x\sqrt{\dfrac{{{m}_{2}}}{{{k}_{eq}}}}=2x\sqrt{\dfrac{8}{2k}} $
Now let’s substitute equation (1) in the above equation
 $ T=2x\sqrt{\dfrac{8}{8{{x}^{2}}}}=2 $
 $ \Rightarrow f=\dfrac{1}{2}Hz $ .
Hence the correct option is (D).

Note: Spring constant is defined as the ratio of force needed to compress or expand the spring to the distance of expansion or contraction of the spring. It is used to demonstrate the stability or instability of the spring. The S.I. unit of spring constant is Newton per metre. After solving for T do not forget to calculate f. If you do then you might make a mistake in choosing option (c).