The equation: \(Y=A\sin(\omega t+\phi_0)\) represents the time-displacement relation of simple harmonic motion (SHM). At \(t=0,\) the displacement of the particle is \(Y=\dfrac{A}{2},\) and it is moving in the negative \(x\text-\)direction. The initial phase angle \(\phi_0\)​ is:
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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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 +\(A \over 2\) and is moving towards \(x = +A\), then \(\delta=\)
1. \({\pi \over 3 }~ \text{rad}\) 2. \({\pi \over 6 }~ \text{rad}\)
3. \({\pi \over 4 } ~\text{rad}\) 4. \({5\pi \over 6 }~\text{rad}\)
Subtopic:  Phasor Diagram |
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