This time period of a particle undergoing SHM is 16s. It starts motion from the mean position. After 2s, velocity is $0.4 {\mathrm{ms}}^{-1}$. The amplitude is:

1. 1.44m

2. 0.72m

3. 2.88m

4. 0.36m

A particle is performing simple harmonic motion along the x-axis with amplitude 4 cm and time period 1.2 s. The minimum time taken by the particle to move from x=+2 cm to x=+4cm and back again is given by:

1. 0.4 sec

2. 0.3 sec

3. 0.2 sec

4. 0.6 sec

The acceleration of a particle performing SHM is $12 \mathrm{cm} {\sec }^{-2}$ at a distance of 3 cm from the mean position. Its time period is:

1. 2.0s

2. 3.14s

3. 0.5s

4. 1.0s

The acceleration ${\mathrm{d}}^2\mathrm{x}/{\mathrm{dt}}^2$ of a particle varies with displacement x as $\frac{d^{2}x}{d{t}^2}=-kx$

where k is a constant of the motion. The time period T of the motion is equal to :

1. $2\mathrm{\pi k}$

2. $2\mathrm{\pi }\sqrt{\mathrm{k}}$

3. $2\mathrm{\pi }/\sqrt{\mathrm{k}}$

4. $2\mathrm{\pi }/\mathrm{k}$

A coin is placed on a horizontal platform, which undergoes horizontal SHM about a mean position O. The coin placed on the platform does not slip, the coefficient of friction between the coin and the platform is $\mathrm{\mu }$. The amplitude of oscillation is gradually increased. The coil will begin to slip on the platform for the first time:

1. at the mean position

2. at the extreme position of oscillations

3. for an amplitude of $\mathrm{\mu g}/{\mathrm{\omega }}^2$

4. for an amplitude of $\mathrm{g}/{\mathrm{\mu \omega }}^2$

A particle moves according to the law, \(x = r \mathrm{cos}\left(\frac{\pi t}{2}\right )\). The distance covered by it in the time interval between \(t=0\) to \(t = 3~\text{s}\) is:
1. \(r\)
2. \(2r\)
3. \(3r\)
4. \(4r\)

A particle is executing SHM of period 24 sec and of amplitude 41 cm with O as equilibrium position. The minimum time in seconds taken by the particle to go from P to Q, where OP=-9cm  and OQ=40cm is:

1. 5

2. 6

3. 7

4. 9

The figure shows the circular motion of a particle which is at the topmost point on the y-axis at t=0. The radius of the circle is B and the sense of revolution is clockwise.  The time period is indicated in the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is:
                   

(1) x(t) = Bsin $\left(\frac{2\mathrm{\pi t}}{30}\right)$

(2) x(t) = Bcos $\left(\frac{\mathrm{\pi t}}{15}\right)$

(3) x(t) = Bsin $\left(\frac{\mathrm{\pi t}}{15}+\frac{\mathrm{\pi }}{2}\right)$

(4) x(t) = Bcos$\left(\frac{\mathrm{\pi t}}{15}+\frac{\mathrm{\pi }}{2}\right)$

A $1.00\times {10}^{-20} kg$ particle is vibrating with simple harmonic motion with a period of $1.00\times {10}^{-5} s$ and a maximum speed of $1.00\times {10}^3 m/s$. The maximum displacement of the particle from the mean position is:

(1) 1.59 mm

(2) 1.00 cm

(3) 10 m

(4) None of these

Which one of the following equations does not represent SHM, x = displacement, and t = time? Parameters a, b and c are the constants of motion.

(1) x = a sin bt

(2) x = a cos bt + c

(3) x = a sin bt + c cos bt

(4) x = a sec bt + c cosec bt