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Q. No.If a particle undergoes simple harmonic motion with displacement:
\(Y=A\sin(\omega t+\phi_0),\)
and at \(t=0\) its displacement is \(Y=\dfrac{A}{2},\) while it is moving in the negative \(x \text-\)direction, what is the initial phase angle \(\phi_0 \text{?}\)
1. \(\dfrac{\pi}{6}\)
2. \(\dfrac{\pi}{3}\)
3. \(\dfrac{5\pi}{6}\)
4. \(\dfrac{2\pi}{3}\)
Subtopic: Phasor Diagram |
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The time period of a simple pendulum is \(T\). The time taken to complete \(\dfrac{5}{8}\) oscillations starting from the mean position is \(\dfrac{\alpha }{\beta}T\). The value of \(\alpha \) is:
| 1. | \(3\) | 2. | \(6\) |
| 3. | \(7\) | 4. | \(10\) |
Subtopic: Phasor Diagram |
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Level 2: 60%+
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The phasor of a particle performing SHM is as shown in the diagram. The SHM has an angular frequency \(\omega\) and at \(t=0\), the phasor lies along \(OP.\) At any time \(t\) further, the projection of the phasor along the \(y \text{-}\)axis is given by:
| 1. | \(R ~\text{sin} \left(\omega t+\dfrac{\pi}{6}\right) \) | 2. | \(R~ \text{cos} \left(\omega t+\dfrac{\pi}{6}\right) \) |
| 3. | \(R~ \text{sin} \left(\omega t-\dfrac{\pi}{6}\right) \) | 4. | \(R~ \text{cos} \left(\omega t-\dfrac{\pi}{6}\right) \) |
Subtopic: Phasor Diagram |
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For a particle performing linear SHM, its position \((x)\) as a function of time \((t)\) is given by \(x = A \sin (\omega t + \delta)\). If, at \(t=0\), particle is at +\(\dfrac{A}{2}\) and is moving towards \(x = +A \), then \(\delta=\)
| 1. | \(\dfrac{\pi}{3}~ \text{rad}\) | 2. | \(\dfrac{\pi}{6}~ \text{rad}\) |
| 3. | \(\dfrac{\pi}{4} ~\text{rad}\) | 4. | \(\dfrac{5\pi}{6}~\text{rad}\) |
Subtopic: Phasor Diagram |
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Q. No.
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