For a particle performing liner SHM, displacement varies with time as `x=7 sin (omega t)+24 cos (omega t)` Amplitude of oscillation of particle is

Answer

To find the amplitude of oscillation for the particle performing linear simple harmonic motion (SHM) given by the displacement equation \( x = 7 \sin(\omega t) + 24 \cos(\omega t) \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given equation can be compared to the standard form of SHM:
\[ x = a \sin(\omega t) + b \cos(\omega t) \]
where \( a = 7 \) and \( b = 24 \).

2. **Use the amplitude formula**:
The amplitude \( A \) of the oscillation can be calculated using the formula:
\[ A = \sqrt{a^2 + b^2} \]

3. **Calculate \( a^2 \) and \( b^2 \)**:
\[ a^2 = 7^2 = 49 \]
\[ b^2 = 24^2 = 576 \]

4. **Sum \( a^2 \) and \( b^2 \)**:
\[ a^2 + b^2 = 49 + 576 = 625 \]

5. **Take the square root to find the amplitude**:
\[ A = \sqrt{625} = 25 \]

6. **State the final answer**:
The amplitude of oscillation of the particle is \( 25 \) units.

### Final Answer:
The amplitude of oscillation of the particle is \( 25 \) units.