Displacement-time equation of a particle executing SHM is x=4sin(`omega t) + 3sin(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 by the equation:

\[ x = 4 \sin(\omega t) + 3 \sin(\omega t + \frac{\pi}{3}) \]

we will follow these steps:

### Step 1: Identify the components of SHM
The equation consists of two sine functions:
- The first term \( 4 \sin(\omega t) \) has an amplitude of 4 cm.
- The second term \( 3 \sin(\omega t + \frac{\pi}{3}) \) has an amplitude of 3 cm and a phase shift of \( \frac{\pi}{3} \).

### Step 2: Represent the components as vectors
We can represent these two components as vectors in a phasor diagram:
- The vector for the first component \( 4 \sin(\omega t) \) is represented as a vector of length 4 at an angle of 0 degrees (along the x-axis).
- The vector for the second component \( 3 \sin(\omega t + \frac{\pi}{3}) \) is represented as a vector of length 3 at an angle of \( \frac{\pi}{3} \) radians (or 60 degrees).

### Step 3: Use the parallelogram law of vector addition
To find the resultant amplitude \( A \), we can use the formula for the resultant of two vectors:

\[ A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)} \]

where:
- \( A_1 = 4 \) (amplitude of the first component)
- \( A_2 = 3 \) (amplitude of the second component)
- \( \phi = \frac{\pi}{3} \) (phase difference between the two components)

### Step 4: Calculate \( A \)
Substituting the values into the formula:

\[ A = \sqrt{4^2 + 3^2 + 2 \cdot 4 \cdot 3 \cdot \cos\left(\frac{\pi}{3}\right)} \]

Calculating each term:
- \( 4^2 = 16 \)
- \( 3^2 = 9 \)
- \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \)

Now substituting these values:

\[ A = \sqrt{16 + 9 + 2 \cdot 4 \cdot 3 \cdot \frac{1}{2}} \]
\[ = \sqrt{16 + 9 + 12} \]
\[ = \sqrt{37} \]

### Step 5: Approximate the result
Now we can approximate \( \sqrt{37} \). The square root of 37 is approximately 6.08 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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