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. \( \sqrt{\dfrac{m}{4 \pi \varepsilon_0 ~qQ} }\)
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 \(\mathrm{x\text-}\)projection of the radius vector of the rotating particle \(P\) will be:
1. \(x \left( t \right) = B\) \(\text{sin} \left(\dfrac{2 πt}{30}\right)\)
2. \(x \left( t \right) = B\) \(\text{cos} \left(\dfrac{πt}{15}\right)\)
3. \(x \left( t \right) = B\) \(\text{sin} \left(\dfrac{πt}{15} + \dfrac{\pi}{2}\right)\)
4. \(x \left( t \right) = B\) \(\text{cos} \left(\dfrac{πt}{15} + \dfrac{\pi}{2}\right)\)
1. | Phase of the oscillator is the same at \(t = 0~\text{s}~\text{and}~t = 2~\text{s}\). |
2. | Phase of the oscillator is the same at \(t = 2~\text{s}~\text{and}~t = 6~\text{s}\). |
3. | Phase of the oscillator is the same at \(t = 1~\text{s}~\text{and}~t = 7~\text{s}\). |
4. | Phase of the oscillator is the same at \(t = 1~\text{s}~\text{and}~t = 5~\text{s}\). |
1. | \(1,2~\text{and}~4\) | 2. | \(1~\text{and}~3\) |
3. | \(2~\text{and}~4\) | 4. | \(3~\text{and}~4\) |
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 |
A spring having a spring constant of \(1200\) N/m is mounted on a horizontal table as shown in the figure. A mass of \(3\) kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of \(2.0\) cm and released. The frequency of oscillations will be:
1. | \(3.0~\text{s}^{-1}\) | 2. | \(2.7~\text{s}^{-1}\) |
3. | \(1.2~\text{s}^{-1}\) | 4. | \(3.2~\text{s}^{-1}\) |
1. | \(0.01~\text{Hz}\) | 2. | \(0.02~\text{Hz}\) |
3. | \(0.03~\text{Hz}\) | 4. | \(0.04~\text{Hz}\) |
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. \(l_A = \frac{l_B}{4}\)
2. \(l_A= 4l_B\)
3. \(l_A= 2l_B~\&~M_A=2M_B\)
4. \(l_A= \frac{l_B}{2}~\&~M_A=\frac{M_B}{2}\)
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