Displacement-time equation of a particle executing SHM is x=4sin`omega`t+3tsin(omegat+pi//3)`
Here, x is in cm and t ini sec. The amplitude of oscillation of the particle is approximately.

Options

A. 7cm
B. 5cm
C. 6cm
D. 9cm

Answer

To find the amplitude of oscillation of the particle executing simple harmonic motion (SHM) given the displacement-time equation \( x = 4 \sin(\omega t) + 3 \sin(\omega t + \frac{\pi}{3}) \), we can follow these steps:

### Step 1: Expand the second term
We start with the equation:
\[ x = 4 \sin(\omega t) + 3 \sin(\omega t + \frac{\pi}{3}) \]
Using the sine addition formula, we can expand the second term:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
Thus,
\[ \sin(\omega t + \frac{\pi}{3}) = \sin(\omega t) \cos\left(\frac{\pi}{3}\right) + \cos(\omega t) \sin\left(\frac{\pi}{3}\right) \]
Substituting the values of \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\) and \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\), we have:
\[ \sin(\omega t + \frac{\pi}{3}) = \sin(\omega t) \cdot \frac{1}{2} + \cos(\omega t) \cdot \frac{\sqrt{3}}{2} \]
Now substituting this back into the equation for \(x\):
\[ x = 4 \sin(\omega t) + 3 \left(\frac{1}{2} \sin(\omega t) + \frac{\sqrt{3}}{2} \cos(\omega t)\right) \]
This simplifies to:
\[ x = 4 \sin(\omega t) + \frac{3}{2} \sin(\omega t) + \frac{3\sqrt{3}}{2} \cos(\omega t) \]
Combining the sine terms:
\[ x = \left(4 + \frac{3}{2}\right) \sin(\omega t) + \frac{3\sqrt{3}}{2} \cos(\omega t) \]
\[ x = \frac{8 + 3}{2} \sin(\omega t) + \frac{3\sqrt{3}}{2} \cos(\omega t) \]
\[ x = \frac{11}{2} \sin(\omega t) + \frac{3\sqrt{3}}{2} \cos(\omega t) \]

### Step 2: Identify the amplitude
The general form of the SHM equation is:
\[ x = A \sin(\omega t + \phi) \]
where \(A\) is the amplitude. To find \(A\), we can use the formula:
\[ A = \sqrt{(B^2 + C^2)} \]
where \(B\) is the coefficient of \(\sin(\omega t)\) and \(C\) is the coefficient of \(\cos(\omega t)\).

Here, \(B = \frac{11}{2}\) and \(C = \frac{3\sqrt{3}}{2}\).

Calculating \(A\):
\[ A = \sqrt{\left(\frac{11}{2}\right)^2 + \left(\frac{3\sqrt{3}}{2}\right)^2} \]
\[ = \sqrt{\frac{121}{4} + \frac{27}{4}} = \sqrt{\frac{148}{4}} = \sqrt{37} \]

### Step 3: Approximate the amplitude
Calculating \(\sqrt{37}\):
\[ \sqrt{37} \approx 6.08 \text{ cm} \]
Thus, the amplitude of oscillation of the particle is approximately:
\[ \text{Amplitude} \approx 6 \text{ cm} \]

### Final Answer
The amplitude of oscillation of the particle is approximately **6 cm**.

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