If the phase angle for a block -spring system in SHM is `pi//8` rad and the block's position is given by `x= x_(m) cos(omega t + phi)`, what is the ratio of the kinetic energy to the potential energy at time t= 0?

Answer

To find the ratio of kinetic energy (KE) to potential energy (PE) for a block-spring system in simple harmonic motion (SHM) at time \( t = 0 \), we will follow these steps:

### Step 1: Write the expressions for kinetic energy and potential energy

The kinetic energy \( KE \) in SHM is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
where \( v \) is the velocity of the block.

The potential energy \( PE \) in SHM is given by the formula:
\[ PE = \frac{1}{2} k x^2 \]
where \( x \) is the displacement of the block from the equilibrium position.

### Step 2: Determine the expressions for velocity and displacement at \( t = 0 \)

The position of the block is given by:
\[ x = x_m \cos(\omega t + \phi) \]
At \( t = 0 \):
\[ x = x_m \cos(\phi) \]
Substituting \( \phi = \frac{\pi}{8} \):
\[ x = x_m \cos\left(\frac{\pi}{8}\right) \]

The velocity \( v \) can be found by differentiating the position with respect to time:
\[ v = \frac{dx}{dt} = -x_m \omega \sin(\omega t + \phi) \]
At \( t = 0 \):
\[ v = -x_m \omega \sin(\phi) = -x_m \omega \sin\left(\frac{\pi}{8}\right) \]

### Step 3: Substitute values into the kinetic and potential energy formulas

Now we can substitute the values of \( x \) and \( v \) into the formulas for kinetic and potential energy.

1. **Kinetic Energy at \( t = 0 \)**:
\[ KE = \frac{1}{2} m \left(-x_m \omega \sin\left(\frac{\pi}{8}\right)\right)^2 = \frac{1}{2} m x_m^2 \omega^2 \sin^2\left(\frac{\pi}{8}\right) \]

2. **Potential Energy at \( t = 0 \)**:
\[ PE = \frac{1}{2} k \left(x_m \cos\left(\frac{\pi}{8}\right)\right)^2 = \frac{1}{2} k x_m^2 \cos^2\left(\frac{\pi}{8}\right) \]

### Step 4: Find the ratio of kinetic energy to potential energy

Now we can find the ratio \( \frac{KE}{PE} \):
\[ \frac{KE}{PE} = \frac{\frac{1}{2} m x_m^2 \omega^2 \sin^2\left(\frac{\pi}{8}\right)}{\frac{1}{2} k x_m^2 \cos^2\left(\frac{\pi}{8}\right)} \]
Canceling \( \frac{1}{2} \) and \( x_m^2 \):
\[ \frac{KE}{PE} = \frac{m \omega^2 \sin^2\left(\frac{\pi}{8}\right)}{k \cos^2\left(\frac{\pi}{8}\right)} \]

### Step 5: Substitute \( \omega^2 \) using \( \omega = \sqrt{\frac{k}{m}} \)

Substituting \( \omega^2 = \frac{k}{m} \):
\[ \frac{KE}{PE} = \frac{m \left(\frac{k}{m}\right) \sin^2\left(\frac{\pi}{8}\right)}{k \cos^2\left(\frac{\pi}{8}\right)} = \frac{\sin^2\left(\frac{\pi}{8}\right)}{\cos^2\left(\frac{\pi}{8}\right)} = \tan^2\left(\frac{\pi}{8}\right) \]

### Step 6: Calculate the value of \( \tan^2\left(\frac{\pi}{8}\right) \)

Using a calculator or trigonometric tables, we find:
\[ \tan\left(\frac{\pi}{8}\right) \approx 0.414 \]
Thus:
\[ \tan^2\left(\frac{\pi}{8}\right) \approx (0.414)^2 \approx 0.171 \]

### Final Answer
The ratio of kinetic energy to potential energy at \( t = 0 \) is approximately:
\[ \frac{KE}{PE} \approx 0.171 \]