The SHM of a particle is given by the equation `x=2 sin omega t + 4 cos omega t`. Its amplitude of oscillation is

Options

A. 4 units
B. 2 units
C. 6 units
D. `2sqrt5` units

Answer

To find the amplitude of the oscillation given the equation of simple harmonic motion (SHM) \( x = 2 \sin(\omega t) + 4 \cos(\omega t) \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components**:
The given equation can be rewritten in terms of sine and cosine components:
\[ x = 2 \sin(\omega t) + 4 \cos(\omega t) \]
Here, the coefficients of \(\sin(\omega t)\) and \(\cos(\omega t)\) are 2 and 4, respectively.

2. **Use the Phasor Representation**:
We can represent each term as a vector (phasor) in a coordinate system:
- The vector for \(2 \sin(\omega t)\) has a magnitude of 2 and is directed along the y-axis.
- The vector for \(4 \cos(\omega t)\) has a magnitude of 4 and is directed along the x-axis.

3. **Calculate the Resultant Amplitude**:
To find the resultant amplitude \(R\), we use the Pythagorean theorem since the two vectors are perpendicular to each other:
\[ R = \sqrt{(4)^2 + (2)^2} \]
\[ R = \sqrt{16 + 4} = \sqrt{20} \]

4. **Simplify the Result**:
The square root of 20 can be simplified:
\[ R = \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \]

5. **Final Answer**:
Therefore, the amplitude of the oscillation is:
\[ R = 2\sqrt{5} \text{ meters} \]

### Summary of the Solution:
The amplitude of the oscillation given by the equation \( x = 2 \sin(\omega t) + 4 \cos(\omega t) \) is \( 2\sqrt{5} \) meters.