A particle executes SHM with a frequency of 25 vib/s and has an amplitude of 0.02m. The speed of the particle 0.30 seconds after it is in the equilibrium position is (n`pi`)m/s. Find value of n

Answer

To solve the problem step by step, we will use the concepts of Simple Harmonic Motion (SHM).

### Step 1: Understand the parameters given
- Frequency (f) = 25 vibrations per second (Hz)
- Amplitude (A) = 0.02 m
- Time (t) = 0.30 seconds

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) can be calculated using the formula:
\[ \omega = 2\pi f \]
Substituting the value of frequency:
\[ \omega = 2\pi \times 25 = 50\pi \, \text{rad/s} \]

### Step 3: Write the equation of motion for SHM
The displacement (x) of a particle in SHM can be expressed as:
\[ x = A \sin(\omega t) \]
Substituting the values of A and ω:
\[ x = 0.02 \sin(50\pi t) \]

### Step 4: Calculate the displacement at t = 0.30 seconds
Substituting t = 0.30 seconds into the equation:
\[ x = 0.02 \sin(50\pi \times 0.30) \]
Calculating the argument of the sine function:
\[ 50\pi \times 0.30 = 15\pi \]
Thus, the displacement becomes:
\[ x = 0.02 \sin(15\pi) \]
Since \(\sin(15\pi) = 0\):
\[ x = 0.02 \times 0 = 0 \]

### Step 5: Calculate the velocity
The velocity (v) in SHM is given by the formula:
\[ v = \omega A \cos(\omega t) \]
Substituting the values:
\[ v = (50\pi)(0.02) \cos(50\pi \times 0.30) \]
Calculating the cosine term:
\[ \cos(15\pi) = -1 \]
Thus, the velocity becomes:
\[ v = (50\pi)(0.02)(-1) = -1\pi \, \text{m/s} \]

### Step 6: Express the velocity in terms of n
The problem states that the velocity can be expressed as \(n\pi\) m/s. From our calculation:
\[ v = -1\pi \, \text{m/s} \]
This means:
\[ n = -1 \]

### Final Answer
The value of \(n\) is \(-1\).