Average value of KE and PE over entire time period is

Options

A. `0,1/2momega^2A^2`
B. `1/2momega^2A^2,0`
C. `1/2momega^2A^2,1/2momega^2A^2`
D. `1/4momega^2A^2,1/4momega^2A^2`

Answer

To find the average values of kinetic energy (KE) and potential energy (PE) over one complete cycle of a simple harmonic motion, we can follow these steps:

### Step 1: Write the expressions for KE and PE
The kinetic energy (KE) at any instant is given by:
\[ KE = \frac{1}{2} m v^2 \]
The potential energy (PE) at any instant is given by:
\[ PE = \frac{1}{2} k x^2 \]

### Step 2: Express the displacement and velocity
In simple harmonic motion, the displacement \(x\) can be expressed as:
\[ x = A \sin(\omega t) \]
where \(A\) is the amplitude and \(\omega\) is the angular frequency.

To find the velocity \(v\), we differentiate \(x\) with respect to time \(t\):
\[ v = \frac{dx}{dt} = \omega A \cos(\omega t) \]

### Step 3: Substitute \(v\) into the KE expression
Now, substituting \(v\) into the kinetic energy expression:
\[ KE = \frac{1}{2} m (\omega A \cos(\omega t))^2 = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t) \]

### Step 4: Substitute \(x\) into the PE expression
Next, substituting \(x\) into the potential energy expression:
\[ PE = \frac{1}{2} k (A \sin(\omega t))^2 = \frac{1}{2} k A^2 \sin^2(\omega t) \]
Since \(k = m \omega^2\), we can write:
\[ PE = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t) \]

### Step 5: Calculate the average values over one complete cycle
The average value of a function \(f(t)\) over one complete cycle (period \(T\)) is given by:
\[ \text{Average} = \frac{1}{T} \int_0^T f(t) dt \]

For kinetic energy:
\[ \text{Average KE} = \frac{1}{T} \int_0^T \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t) dt \]
Using the identity \(\int_0^T \cos^2(\omega t) dt = \frac{T}{2}\), we find:
\[ \text{Average KE} = \frac{1}{2} m \omega^2 A^2 \cdot \frac{1}{2} = \frac{1}{4} m \omega^2 A^2 \]

For potential energy:
\[ \text{Average PE} = \frac{1}{T} \int_0^T \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t) dt \]
Similarly, using the identity \(\int_0^T \sin^2(\omega t) dt = \frac{T}{2}\), we find:
\[ \text{Average PE} = \frac{1}{2} m \omega^2 A^2 \cdot \frac{1}{2} = \frac{1}{4} m \omega^2 A^2 \]

### Final Result
Thus, the average values of kinetic energy and potential energy over one complete cycle are:
\[ \text{Average KE} = \frac{1}{4} m \omega^2 A^2 \]
\[ \text{Average PE} = \frac{1}{4} m \omega^2 A^2 \]