Question
A body vibrates in simple harmoic motion with a frequency of 50Hz and an amplitude of 4cm. Find the acceleration at the middle and end of the path of oscillation
Answers
GPT-4o mini
In simple harmonic motion (SHM), the acceleration of the oscillating body can be calculated using the formula:
\[
a = -\omega^2 x
\]
where:
- \( a \) is the acceleration,
- \( \omega \) is the angular frequency, and
- \( x \) is the displacement from the equilibrium position.
1. **Calculate the angular frequency (\(\omega\))**:
The angular frequency is related to the frequency (\(f\)) by the formula:
\[
\omega = 2\pi f
\]
Given that \(f = 50 \, \text{Hz}\),
\[
\omega = 2\pi \times 50 \approx 314.16 \, \text{rad/s}
\]
2. **Acceleration at the middle of the path**:
At the middle of the path of oscillation, the displacement \(x = 0\):
\[
a = -\omega^2 x = -\omega^2 \cdot 0 = 0 \, \text{m/s}^2
\]
Therefore, the acceleration at the middle of the path is \(0 \, \text{m/s}^2\).
3. **Acceleration at the end of the path**:
At the end of the path of oscillation, the displacement \(x\) is at its maximum, which is equal to the amplitude:
\[
x = A = 4 \, \text{cm} = 0.04 \, \text{m}
\]
Now, substituting the values into the acceleration formula:
\[
a = -\omega^2 x = -\omega^2 \cdot 0.04
\]
\[
a = - (314.16)^2 \cdot 0.04
\]
\[
a = - 9873.68 \cdot 0.04 \approx -395.07 \, \text{m/s}^2
\]
The negative sign indicates that the acceleration is directed towards the equilibrium position.
**Summary**:
- The acceleration at the middle of the path: \(0 \, \text{m/s}^2\)
- The acceleration at the end of the path: approximately \(-395.07 \, \text{m/s}^2\) (which indicates it is directed towards the equilibrium position).
\[
a = -\omega^2 x
\]
where:
- \( a \) is the acceleration,
- \( \omega \) is the angular frequency, and
- \( x \) is the displacement from the equilibrium position.
1. **Calculate the angular frequency (\(\omega\))**:
The angular frequency is related to the frequency (\(f\)) by the formula:
\[
\omega = 2\pi f
\]
Given that \(f = 50 \, \text{Hz}\),
\[
\omega = 2\pi \times 50 \approx 314.16 \, \text{rad/s}
\]
2. **Acceleration at the middle of the path**:
At the middle of the path of oscillation, the displacement \(x = 0\):
\[
a = -\omega^2 x = -\omega^2 \cdot 0 = 0 \, \text{m/s}^2
\]
Therefore, the acceleration at the middle of the path is \(0 \, \text{m/s}^2\).
3. **Acceleration at the end of the path**:
At the end of the path of oscillation, the displacement \(x\) is at its maximum, which is equal to the amplitude:
\[
x = A = 4 \, \text{cm} = 0.04 \, \text{m}
\]
Now, substituting the values into the acceleration formula:
\[
a = -\omega^2 x = -\omega^2 \cdot 0.04
\]
\[
a = - (314.16)^2 \cdot 0.04
\]
\[
a = - 9873.68 \cdot 0.04 \approx -395.07 \, \text{m/s}^2
\]
The negative sign indicates that the acceleration is directed towards the equilibrium position.
**Summary**:
- The acceleration at the middle of the path: \(0 \, \text{m/s}^2\)
- The acceleration at the end of the path: approximately \(-395.07 \, \text{m/s}^2\) (which indicates it is directed towards the equilibrium position).