The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm s^-1 and 50 cms^-2. Find the position of the particle when the speed is 8cms^-1.

Explanation

To solve the problem step by step, we will follow the principles of simple harmonic motion (SHM) and the equations governing it.

### Step 1: Identify the given values
- Maximum speed (\(v_{max}\)) = 10 cm/s
- Maximum acceleration (\(a_{max}\)) = 50 cm/s²
- Speed at which we need to find the position (\(v\)) = 8 cm/s

### Step 2: Write the equations for maximum speed and maximum acceleration
1. The maximum speed in SHM is given by:
\[ v_{max} = \omega A \]
where \(A\) is the amplitude and \(\omega\) is the angular frequency.

2. The maximum acceleration in SHM is given by:
\[ a_{max} = \omega^2 A \]

### Step 3: Relate the two equations
From the two equations, we can express \(\omega\) in terms of \(a_{max}\) and \(v_{max}\):
- From the first equation:
\[ \omega = \frac{v_{max}}{A} \]
- From the second equation:
\[ \omega^2 = \frac{a_{max}}{A} \]

### Step 4: Divide the two equations
Dividing the second equation by the first gives:
\[ \frac{a_{max}}{v_{max}} = \frac{\omega^2 A}{\omega A} \]
This simplifies to:
\[ \frac{a_{max}}{v_{max}} = \omega \]

### Step 5: Calculate \(\omega\)
Substituting the known values:
\[ \omega = \frac{50 \, \text{cm/s}^2}{10 \, \text{cm/s}} = 5 \, \text{s}^{-1} \]

### Step 6: Calculate the amplitude \(A\)
Now, substituting \(\omega\) back into the equation for maximum speed:
\[ 10 = 5A \implies A = \frac{10}{5} = 2 \, \text{cm} \]

### Step 7: Use the velocity equation to find position \(x\)
The velocity in SHM is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
Substituting the known values:
\[ 8 = 5 \sqrt{2^2 - x^2} \]
This simplifies to:
\[ 8 = 5 \sqrt{4 - x^2} \]

### Step 8: Solve for \(x\)
Dividing both sides by 5:
\[ \frac{8}{5} = \sqrt{4 - x^2} \]
Squaring both sides gives:
\[ \left(\frac{8}{5}\right)^2 = 4 - x^2 \]
Calculating \(\left(\frac{8}{5}\right)^2\):
\[ \frac{64}{25} = 4 - x^2 \]
Rearranging gives:
\[ x^2 = 4 - \frac{64}{25} \]
Converting 4 to a fraction:
\[ 4 = \frac{100}{25} \implies x^2 = \frac{100}{25} - \frac{64}{25} = \frac{36}{25} \]
Taking the square root:
\[ x = \pm \frac{6}{5} \, \text{cm} \]

### Final Answer
The position of the particle when the speed is 8 cm/s is:
\[ x = \pm 1.2 \, \text{cm} \]