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

The bob of a simple pendulum of length L is released at time t = 0 from a position of small angular displacement. Its linear displacement at time t is given by :

(1) $X=a \sin 2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}\times \mathrm{t}$

(2) $X=a \cos 2\mathrm{\pi }\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\times \mathrm{t}$

(3) $X=a \sin \sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\times \mathrm{t}$

(4) $X=a \cos \sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\times \mathrm{t}$

A particle in SHM is described by the displacement function $x\left(t\right)=A \cos (\omega t+\varphi ), \omega =2\pi /T.$ If the initial (t = 0) position of the particle is 1 cm, its initial velocity is $\pi cm s^{-1}$ and its angular frequency is $\pi s^{-1}$, then the amplitude of its motion is:

(1) $\pi cm$

(2) 2 cm

(3) $\sqrt{2} cm$

(4) 1 cm

A particle starts SHM from the mean position. Its amplitude is 'a' and total energy E. At one instant its kinetic energy is 3E/4. Its displacement at this instant is :

(1) $y=a/\sqrt{2}$

(2) $y=\frac{a}{2}$

(3) $y=\frac{a}{\sqrt{3/2}}$

(4) y = a

A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position, is its energy half potential and half kinetic?

(1) 1 cm

(2) $\sqrt{2} cm$

(3) 3 cm

(4) $2\sqrt{2} cm$