The displacement of a SHO is given by y = 2 sin `(2pit+pi//4)m`. The ratio of its initial displacement to maximum displacement is

Options

A. `1//sqrt(2)`
B. `sqrt(2)//1`
C. 2
D. zero

Explanation

To solve the problem, we need to find the ratio of the initial displacement to the maximum displacement of a simple harmonic oscillator (SHO) given the displacement equation:

**Step 1: Identify the displacement equation.**
The displacement of the SHO is given by:
\[ y = 2 \sin(2\pi t + \frac{\pi}{4}) \]

**Step 2: Calculate the initial displacement.**
To find the initial displacement \( y_0 \), we substitute \( t = 0 \) into the displacement equation:
\[ y_0 = 2 \sin(2\pi(0) + \frac{\pi}{4}) \]
\[ y_0 = 2 \sin(\frac{\pi}{4}) \]

**Step 3: Evaluate \( \sin(\frac{\pi}{4}) \).**
The value of \( \sin(\frac{\pi}{4}) \) is:
\[ \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \]
Thus, we have:
\[ y_0 = 2 \cdot \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

**Step 4: Determine the maximum displacement.**
The maximum displacement \( y_{\text{max}} \) occurs when the sine function reaches its maximum value of 1. Therefore:
\[ y_{\text{max}} = 2 \cdot 1 = 2 \]

**Step 5: Calculate the ratio of initial displacement to maximum displacement.**
Now we can find the ratio of the initial displacement to the maximum displacement:
\[ \text{Ratio} = \frac{y_0}{y_{\text{max}}} = \frac{\sqrt{2}}{2} \]

**Step 6: Simplify the ratio.**
This simplifies to:
\[ \text{Ratio} = \frac{1}{\sqrt{2}} \]

Thus, the final answer is:
\[ \text{Ratio of initial displacement to maximum displacement} = \frac{1}{\sqrt{2}} \]

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