The equation of a simple harmonic oscillator with amplitude 5 cm and period 2 sec is if particle is starting from the mean positions is–

Options

A. `y=5 sin pi t`
B. `y =2 sin 2 pi t //5`
C. `y =5 cos pi t`
D. `y =5 2 pi t //5`

Answer

To find the equation of a simple harmonic oscillator (SHO) with an amplitude of 5 cm and a period of 2 seconds, we can follow these steps:

### Step 1: Identify the parameters
- Amplitude (A) = 5 cm
- Period (T) = 2 seconds

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is related to the period (T) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the given period:
\[ \omega = \frac{2\pi}{2} = \pi \, \text{rad/s} \]

### Step 3: Write the general equation of SHO
The general equation for a simple harmonic oscillator can be expressed as:
\[ y(t) = A \sin(\omega t + \phi) \]
where:
- \(y(t)\) is the displacement at time \(t\),
- \(A\) is the amplitude,
- \(\omega\) is the angular frequency,
- \(\phi\) is the phase constant.

### Step 4: Substitute the known values
Substituting the values of amplitude and angular frequency into the equation:
\[ y(t) = 5 \sin(\pi t + \phi) \]

### Step 5: Determine the phase constant (φ)
Since the particle starts from the mean position at \(t = 0\), we know that \(y(0) = 0\). Thus:
\[ y(0) = 5 \sin(\pi \cdot 0 + \phi) = 0 \]
This implies:
\[ 5 \sin(\phi) = 0 \]
The sine function equals zero when:
\[ \phi = n\pi \quad (n \in \mathbb{Z}) \]
For the simplest case, we can take \(n = 0\), which gives:
\[ \phi = 0 \]

### Step 6: Write the final equation
Substituting \(\phi = 0\) back into the equation gives:
\[ y(t) = 5 \sin(\pi t) \]

### Final Answer:
The equation of the simple harmonic oscillator is:
\[ y(t) = 5 \sin(\pi t) \, \text{cm} \]

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