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. Hint: Use the trigonometric functions to find the equation.
Step 1: Find the horizontal projection of the radius vector.
Let angular velocity of the particle executing circular motion is ω and when it is at Q, itmakes an angle θ as shown in the diagram.
                         
Clearly,                              θ=ωt
Now, we can write              OR=OQcos(90 -θ)
                                            =OQsinθ=OQsinωt
                                            = rsinωt                              [ OQ=r]
Step 2: Find the equation of motion.
                                     x = rsinωt=Bsinωt                [r=B]
                                            =Bsin2πTt=B sin 2π30t
Clearly, this equation represents SHM.