Q. No.A particle is executing a simple harmonic motion. Its maximum acceleration is and maximum velocity is . Then its time period of vibration will be:
1.
\(\frac {\beta^2}{\alpha^2}\)
2.
\(\frac {\beta}{\alpha}\)
3.
\(\frac {\beta^2}{\alpha}\)
4.
\(\frac {2\pi \beta}{\alpha}\)
When two displacements are represented by \(y_1 = a \text{sin}(\omega t)\) and \(y_2 = b\text{cos}(\omega t)\) are superimposed, then the motion is:
| 1. | not simple harmonic. |
| 2. | simple harmonic with amplitude \(\frac{a}{b}\). |
| 3. | simple harmonic with amplitude \(\sqrt{a^2+b^{2}}.\) |
| 4. | simple harmonic with amplitude \(\frac{a+b}{2}\). |
A particle is executing SHM along a straight line. Its velocities at distances \(x_1\) and \(x_2\) from the mean position are \(v_1\) and \(v_2\), respectively. Its time period is:
| 1. | \(2 \pi \sqrt{\dfrac{x_{1}^{2}+x_{2}^{2}}{v_{1}^{2}+v_{2}^{2}}}~\) | 2. | \(2 \pi \sqrt{\dfrac{{x}_{2}^{2}-{x}_{1}^{2}}{{v}_{1}^{2}-{v}_{2}^{2}}}\) |
| 3. | \(2 \pi \sqrt{\dfrac{v_{1}^{2}+v_{2}^{2}}{x_{1}^{2}+x_{2}^{2}}}\) | 4. | \(2 \pi \sqrt{\dfrac{v_{1}^{2}-v_{2}^{2}}{x_{1}^{2}-x_{2}^{2}}}\) |
The oscillation of a body on a smooth horizontal surface is represented by the equation, \(X=A \text{cos}(\omega t)\),
where \(X=\) displacement at time \(t,\) \(\omega=\) frequency of oscillation.
Which one of the following graphs correctly shows the variation of acceleration, \(a\) with time, \(t?\)
(\(T=\) time period) \(a~~O~~T~~t~~\)
| 1. |  |
2. |  |
| 3. |  |
4. |  |
1. Only (IV) does not represent SHM
2. (I) and (III)
3. (I) and (II)
4. Only (I)
A particle of mass \(m\) is released from rest and follows a parabolic path as shown. Assuming that the displacement of the mass from the origin is small, which graph correctly depicts the position of the particle as a function of time?
| 1. |  |
2. |  |
| 3. |  |
4. |  |
1. \(\frac{\pi a\sqrt3}{2T}\)
2. \(\frac{\pi a}{T}\)
3. \(\frac{3\pi^2 a}{T}\)
4. \(\frac{\pi a\sqrt3}{T}\)
1. Acceleration = -k0x + k1x2
2. Acceleration = -k(x+a)
3. Acceleration = k(x+a)
4. Acceleration = kx
1. 1: 10
2. 1: 102
3. 1: 103
4. 1: 104
A point performs simple harmonic oscillation of period \(\mathrm{T}\) and the equation of motion is given by; \(x=a \sin (\omega t+\pi / 6)\). After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity?
1. \( \frac{T}{8} \)
2. \( \frac{T}{6} \)
3. \(\frac{T}{3} \)
4. \( \frac{T}{12}\)
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