Ncert Exercises & Solutions

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-projection of the radius vector of the rotating particle P will be:

1. $x (t) = B$ $\sin (\frac{2 πt}{30})$

2. $x (t) = B$ $\cos (\frac{πt}{15})$

3. $x (t) = B$ $\sin (\frac{πt}{15} + \frac{π}{2})$

4. $x (t) = B$ $\cos (\frac{πt}{15} + \frac{π}{2})$

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

ω

Now, we can write OR=OQcos(90

- θ

)

=OQsin

θ

=OQsin

ω

= rsin

ω

t [

∵

OQ=r]

Step 2: Find the equation of motion.

\Rightarrow

x = rsin

ω

t=Bsin

ω

t [

∵

r=B]

= B \sin \frac{2 π}{T} t = B

\sin

(\frac{2 π}{30} t)

Clearly, this equation represents SHM.

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