A linear SHM is represented by `x=5sqrt(2) ("sin" 2pi t+ "cos" 2pi t)`cm.
What is the amplitude of SHM ?

Options

A. `5sqrt(2)`
B. 50 cm
C. 10 cm
D. 20 cm

Explanation

To find the amplitude of the given linear simple harmonic motion (SHM) represented by the equation:

\[ x = 5\sqrt{2} \sin(2\pi t) + 5\sqrt{2} \cos(2\pi t) \]

we can follow these steps:

### Step 1: Identify the components of the SHM
The equation consists of two terms: one involving sine and the other involving cosine. We can rewrite the equation in a more standard form.

### Step 2: Rewrite the equation
We can factor out the common coefficient \( 5\sqrt{2} \):

\[ x = 5\sqrt{2} (\sin(2\pi t) + \cos(2\pi t)) \]

### Step 3: Use the trigonometric identity
To combine the sine and cosine terms, we can use the trigonometric identity:

\[ A \sin(\omega t) + B \cos(\omega t) = R \sin(\omega t + \phi) \]

where \( R = \sqrt{A^2 + B^2} \) is the resultant amplitude, and \( \phi \) is the phase angle.

In our case, \( A = 5\sqrt{2} \) and \( B = 5\sqrt{2} \).

### Step 4: Calculate the resultant amplitude
Now, we can find \( R \):

\[ R = \sqrt{(5\sqrt{2})^2 + (5\sqrt{2})^2} \]

Calculating the squares:

\[ (5\sqrt{2})^2 = 25 \cdot 2 = 50 \]

So,

\[ R = \sqrt{50 + 50} = \sqrt{100} = 10 \]

### Step 5: State the amplitude
The amplitude of the SHM is therefore:

\[ \text{Amplitude} = 10 \text{ cm} \]

### Final Answer
The amplitude of the SHM is **10 cm**.

---