KCET · Physics · Oscillations
A block of mass \(m\) is connected to a light spring of force constant \(k\). The system is placed inside a damping medium of damping constant \(b\). The instantaneous values of displacement, acceleration and energy of the block are \(x, a\) and \(E\) respectively. The initial amplitude of oscillation is \(A\) and \(\omega^{\prime}\) is the angular frequency of oscillations. The incorrect
expression related to the damped oscillations is
- A \(x=A e^{-\frac{b}{m}} \cos \left(\omega^{\prime} t+\phi\right)\)
- B \(\omega^{\prime}=\sqrt{\frac{k}{m}-\frac{b^2}{4 m^2}}\)
- C \(E=\frac{1}{2} k A^2 e^{-\frac{b t}{m}}\)
- D \(m \frac{d^2 x}{d t^2}+b \frac{d x}{d t}+k x=0\)
Answer & Solution
Correct Answer
(A) \(x=A e^{-\frac{b}{m}} \cos \left(\omega^{\prime} t+\phi\right)\)
Step-by-step Solution
Detailed explanation
Equation of motion for damped oscillation is
\(\frac{m d^2 x}{d t^2}+\frac{b d x}{d t}+k x=0\)
Angular frequency is \(\omega^{\prime}=\sqrt{\frac{k}{m}-\frac{b^2}{4 m^2}}\) Displacement is given by, \(x=A_0 e^{-\frac{b}{2 m} t} \cos (\omega t+\phi)\) and energy is given by \(E=\frac{1}{2} k A^2 e^{-\frac{b}{m} t}\)
\(\frac{m d^2 x}{d t^2}+\frac{b d x}{d t}+k x=0\)
Angular frequency is \(\omega^{\prime}=\sqrt{\frac{k}{m}-\frac{b^2}{4 m^2}}\) Displacement is given by, \(x=A_0 e^{-\frac{b}{2 m} t} \cos (\omega t+\phi)\) and energy is given by \(E=\frac{1}{2} k A^2 e^{-\frac{b}{m} t}\)
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