Question
Question: The mass M shown in the figure oscillates in simple harmonic motion with amplitude A, the amplitude ...
The mass M shown in the figure oscillates in simple harmonic motion with amplitude A, the amplitude of the point P is
A: k2k1A
B: k1k2A
C: k2+k1k1A
D: k2+k1k2A
Solution
Here the mass is oscillating that is undergoing simple harmonic motion and the two springs are attached as shown in the figure. We know whenever the spring is compressed or extended a restoring force comes in to play which pulls it back to its equilibrium position. Both the springs have different spring constants.
Step by step solution:
Since the mass is oscillating and in order to that some external force has to be applied, let it be F.
Now we can write the restoring force develops in the springs as:
F={{k}_{1}}{{x}_{1}} \\\
F={{k}_{2}}{{x}_{2}} \\\
Where x1&x2are the displacement of the first and second spring from their equilibrium position.
⇒x1+x2=A because the maximum displacement is the amplitude of the oscillations.
Writing in terms of Force we get,
\dfrac{{{k}_{2}}A}{{{k}_{2}}+{{k}_{1}}} \\\ F={{k}_{1}}{{x}_{1}} \\\ F={{k}_{2}}{{x}_{2}} \\\ {{x}_{1}}\And {{x}_{2}} \\\ {{x}_{1}}+{{x}_{2}}=A \\\ \Rightarrow A=\dfrac{F}{{{k}_{1}}}+\dfrac{F}{{{k}_{2}}} \\\ \Rightarrow \dfrac{A}{F}=\dfrac{1}{{{k}_{1}}}+\dfrac{1}{{{k}_{2}}} \\\ \Rightarrow F=\dfrac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}A \\\
So, the amplitude of the point: x1=k1F=k1+k2k1A
So, the correct option is (C)
Note: We know spring has a spring of force constant, k and it has the tendency to return to its equilibrium position that is the unstretched position when it is either compressed or stretched. The energy possessed by the spring when it is either stretched or compressed is called its potential energy .
We know potential energy of the spring is given by the formula, U=2kx2, where k is the spring constant of the spring in units of N/m and x is the displacement of the spring from its equilibrium position in units of metres.