Displacement versus time curve for a particle executing SHM is shown in the figure. Choose the correct statement/s.
1. | Phase of the oscillator is the same at t =0 s and t = 2 s. |
2. | Phase of the oscillator is the same at t =2 s and t=6 s. |
3. | Phase of the oscillator is the same at t = 1 s and t=7 s. |
4. | Phase of the oscillator is the same at t=1 s and t=5 s. |
1. | 1, 2 and 4 | 2. | 1 and 3 |
3. | 2 and 4 | 4. | 3 and 4 |
The figure shows the circular motion of a particle. The radius of the circle, the period, the sense of revolution, and the initial position are indicated in the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P will be:
1.
2.
3.
4.
A particle of mass \(m\) and charge \(\text-q\) moves diametrically through a uniformly charged sphere of radius \(R\) with total charge \(Q\). The angular frequency of the particle's simple harmonic motion, if its amplitude \(<R\), is given by:
1. \(\sqrt{\dfrac{qQ}{4 \pi \varepsilon_0 ~mR} }\)
2. \(\sqrt{\dfrac{qQ}{4 \pi \varepsilon_0 ~mR^2} }\)
3. \(\sqrt{\dfrac{qQ}{4 \pi \varepsilon_0 ~mR^3}}\)
4. \( \sqrt{\dfrac{m}{4 \pi \varepsilon_0 ~qQ} }\)
From the given functions, identify the function which represents a periodic motion:
1. | \(e^{\omega t}\) | 2. | \(\text{log}_e(\omega t)\) |
3. | \(\text{sin}\omega t+ \text{cos}\omega t\) | 4. | \(e^{-\omega t}\) |
Identify the correct definition:
1. | If after every certain interval of time, a particle repeats its motion, then the motion is called periodic motion. |
2. | To and fro motion of a particle is called oscillatory motion. |
3. | Oscillatory motion described in terms of single sine and cosine functions is called simple harmonic motion. |
4. | All of the above |
If the time of mean position from amplitude (extreme) position is 6 seconds, then the frequency of SHM will be:
1. | \(0.01\) Hz | 2. | \(0.02\) Hz |
3. | \(0.03\) Hz | 4. | \(0.04\) Hz |
Two spherical bobs of masses \(M_A\) and \(M_B\) are hung vertically from two strings of length \(l_A\) and \(l_B\) respectively. If they are executing SHM with frequency as per the relation \(f_A=2f_B,\) Then:
1.
2.
3.
4.
A spring elongates by a length 'L' when a mass 'M' is suspended to it. Now a tiny mass 'm' is attached to the mass 'M' and then released. The new time period of oscillation will be:
1. \(2 \pi \sqrt{\frac{\left(\right. M + m \left.\right) l}{Mg}}\)
2. \(2 \pi \sqrt{\frac{ml}{Mg}}\)
3. \(2 \pi \sqrt{L / g}\)
4. \(2 \pi \sqrt{\frac{Ml}{\left(\right. m + M \left.\right) g}}\)
The frequency of a simple pendulum in a free-falling lift will be:
1. zero
2. infinite
3. can't say
4. finite
When a mass is suspended separately by two different springs, in successive order, then the time period of oscillations is \(t _1\) and \(t_2\) respectively. If it is connected by both springs as shown in the figure below, then the time period of oscillation becomes \(t_0.\) The correct relation between \(t_0,\) \(t_1\) & \(t_2\) is:
1.
2.
3.
4.