TS EAMCET · Physics · Oscillations
A point mass oscillates along the \(X\)-axis according to the law \(x=x_0 \cos \left(\omega t-\frac{\pi}{4}\right)\). If the acceleration of the particle is written as \(a=A \cos (\omega t-\delta)\), then
- A \(A=x_0 \omega^2, \delta=\frac{-3 \pi}{4}\)
- B \(A=x_0, \delta=-\frac{\pi}{4}\)
- C \(A=x_0 \omega^2, \delta=\frac{\pi}{4}\)
- D \(A=x_0 \omega^2, \delta=\frac{3 \pi}{4}\)
Answer & Solution
Correct Answer
(A) \(A=x_0 \omega^2, \delta=\frac{-3 \pi}{4}\)
Step-by-step Solution
Detailed explanation
Displacement of particle is \(x=x_0 \cos \left(\omega t-\frac{\pi}{4}\right)\) Velocity, \(v=\frac{d x}{d t}\) \(\Rightarrow v=-x_0 \omega \sin \left(\omega t-\frac{\pi}{4}\right)\) Acceleration, \(a=\frac{d v}{d t}=-x_0 \omega^2 \cos \left(\omega t-\frac{\pi}{4}\right)\)…
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