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

Hint: Use the trigonometric functions to find the equation.

Step 1: Find the horizontal projection of the radius vector.

Let the angular velocity of the particle executing circular motion be

ω

$\omega$ and when it is at

Q,

$Q,$ it makes an angle

θ

$\theta$ as shown in the diagram.

Now, we can write

O R = O Q cos (90 - θ)

$OR=OQ\cos(90 - 𝜃 )$

\Rightarrow O R = O Q sin θ = O Q sin ω t

$\Rightarrow OR= OQ\sin 𝜃 =OQ\sin 𝜔 t$

\Rightarrow O R = r sin ω t [O Q = r, θ = ω t]

$\Rightarrow OR= r\sin 𝜔 t~~~~~~~ [OQ=r, \theta = \omega t]$

Step 2: Find the equation of motion.

sin ω t = \frac{x}{B}

$\sin\omega t=\frac{x}{B}$

\Rightarrow x = r sin ω t = B sin ω t

$\Rightarrow x = r\sin 𝜔 t=B\sin 𝜔 t$

$[ ∵ r=B]$
The angular frequency

$(\omega)$ is related to the period by:

$\omega = \frac{2\pi}{T} =\frac{2\pi}{30} ~ \text{rad/s}$

$\Rightarrow x = B\sin 𝜔 t$

$\Rightarrow x = B\sin\left(\frac{2\pi t}{30}\right)$
Hence, option (1) is the correct answer.

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