Q. No.| 1. | \(2 \pi \over K\) | 2. | \(2 \pi K\) |
| 3. | \(2 \pi \over \sqrt{K}\) | 4. | \(2 \pi \sqrt{K}\) |
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)
| 1. |  |
2. |  |
| 3. |  |
4. |  |
The distance covered by a particle undergoing SHM in one time period is: (amplitude \(= A\))
1. zero
2. \(A\)
3. \(2 A\)
4. \(4 A\)
1. \({4a \over T}\)
2. \({2a \over T}\)
3. \({2 \pi \over T}\)
4. \({2a \pi \over T}\)
Then the amplitude of its oscillation is given by:
1. \(A + B\)
2. \(A_{0}+\sqrt{A^{2} + B^{2}} \)
3. \(\sqrt{A^{2} + B^{2}}\)
4. \(\sqrt{A_{0}^{2}+\left( A + B \right)^{2}}\)
| 1. | \(5~\text m, 2~\text s\) | 2. | \(5~\text {cm}, 1~\text s\) |
| 3. | \(5~\text m, 1~\text s\) | 4. | \(5~\text {cm}, 2~\text s\) |
A particle is executing a simple harmonic motion. Its maximum acceleration is \(\alpha\) and maximum velocity is \(\beta.\) Then its time period of vibration will be:
1. \(\dfrac {\beta^2}{\alpha^2}\)
2. \(\dfrac {\beta}{\alpha}\)
3. \(\dfrac {\beta^2}{\alpha}\)
4. \(\dfrac {2\pi \beta}{\alpha}\)
| 1. | \(A_1 \omega_1=A_2 \omega_2=A_3 \omega_3\) |
| 2. | \(A_1 \omega_1^2=A_2 \omega_2^2=A_3 \omega_3^2\) |
| 3. | \(A_1^2 \omega_1=A_2^2 \omega_2=A_3^2 \omega_3\) |
| 4. | \(A_1^2 \omega_1^2=A_2^2 \omega_2^2=A^2\) |
The average velocity of a particle executing SHM in one complete vibration is:
1. zero
2. \(\dfrac{A \omega}{2}\)
3. \(A \omega\)
4. \(\dfrac{A \left(\omega\right)^{2}}{2}\)
| 1. | \(0.01~\text{Hz}\) | 2. | \(0.02~\text{Hz}\) |
| 3. | \(0.03~\text{Hz}\) | 4. | \(0.04~\text{Hz}\) |
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