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 \(\omega \) and when it is at \(Q,\) it makes an angle \(\theta\) as shown in the diagram.

Now, we can write
\( OR=OQ\cos(90 - 𝜃 )\)

\(\Rightarrow OR= OQ\sin 𝜃 =OQ\sin 𝜔 t\)

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

Step 2: Find the equation of motion.
\(\sin\omega t=\frac{x}{B}\)
\(\Rightarrow x = r\sin 𝜔 t=B\sin 𝜔 t \) \( [ ∵ r=B]\)
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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