The displacement of a body executing SHM is given by x= A sin (2$\mathrm{\pi }$t + $\mathrm{\pi }$/3). The first time from t = 0 when the velocity is maximum is :
1. 0.33 sec
2. 0.16 sec
3. 0.25 sec
4. 0.5 sec

The time period of a particle executing SHM is 8 sec. At t = 0 it is at the mean position. The ratio of the distance covered by the particle in the 1st second to the 2nd second is :

1. $\frac{1}{\sqrt{2}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>1}$

2. $\sqrt{2}$

3. $\frac{1}{\sqrt{2}<mspace\; width="1em"></mspace>}$

4. $\sqrt{2}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>1$

Two particles are in SHM in a straight line. Amplitude A and time period T of both the particles are equal. At time t = 0, one particle is at displacement Y₁ = + A and the other at Y₂ = -A/2, and they are approaching towards each other. After what time they cross each other?
1. T/3
2. T/4
3. 5T/6
4. T/6

Two particles execute SHM of the same amplitude of \(20\) cm with the same time period along the same line about the same equilibrium position. The maximum distance between the two is \(20\) cm. Their phase difference in radians is:
1. \(\frac{2\pi}{3}\)
2. \(\frac{\pi}{2}\)
3. \(\frac{\pi}{3}\)
4. \(\frac{\pi}{4}\)

Two particles P and Q describe simple harmonic motions of the same period, the same amplitude, along the same line about the same equilibrium position O. When P and Q are on opposite sides of O at the same distance from O, they have the same speed of 1.2 m/s in the same direction. When their displacements are the same, they have the same speed of 1.6 m/s in opposite directions. The maximum velocity in m/s of either particle is 
1. 2.8
2. 2.5
3. 2.4
4. 2

A pendulum consisting of a small sphere of mass m suspended by an inextensible and massless string of length l is made to swing in a vertical plane. If the breaking strength of the string is 2mg, then the maximum angular amplitude of the displacement from the vertical can be :-

1.  0$°$

2.  30$°$

3.  60$°$

4.  90$°$

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 particle executes S.H.M with amplitude A. If the time taken by the particle to travel from -A to A/2 is 4 seconds, its time period is 

1.  4s

2.  8s 

3.  12 s

4.  18 s

The displacement \( x\) of a particle varies with time \(t\) as \(x = A sin\left (\frac{2\pi t}{T} +\frac{\pi}{3} \right)\)$\mathrm{}$. The time taken by the particle to reach from \(x = \frac{A}{2} \) to \(x = -\frac{A}{2} \) will be:

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 \({x\text-}\)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. \(x \left( t \right) = B\text{cos} \left(\dfrac{πt}{15} + \dfrac{\pi}{2}\right)\)