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

### Step 1: Identify the components of the SHM equation
The equation is given as:
\[ x = 2 \sin(\omega t) + 4 \cos(\omega t) \]
Here, we can identify the coefficients of the sine and cosine functions:
- \( A_1 = 2 \) (coefficient of \( \sin(\omega t) \))
- \( A_2 = 4 \) (coefficient of \( \cos(\omega t) \))

### Step 2: Use the formula for the amplitude of SHM
The amplitude \( A \) of a SHM represented in the form \( x = A_1 \sin(\omega t) + A_2 \cos(\omega t) \) can be calculated using the formula:
\[ A = \sqrt{A_1^2 + A_2^2} \]

### Step 3: Substitute the values of \( A_1 \) and \( A_2 \)
Now, substituting the values of \( A_1 \) and \( A_2 \) into the amplitude formula:
\[ A = \sqrt{(2)^2 + (4)^2} \]

### Step 4: Calculate the squares
Calculating the squares:
\[ A = \sqrt{4 + 16} \]

### Step 5: Sum the squares
Now, sum the squares:
\[ A = \sqrt{20} \]

### Step 6: Simplify the square root
Finally, simplify the square root:
\[ A = \sqrt{4 \times 5} = 2\sqrt{5} \]

### Conclusion
Thus, the amplitude of oscillation is:
\[ \boxed{2\sqrt{5}} \]

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