Question
A particle executes simple harmonic motion
with an amplitude of 2.7 cm.
At what positive displacement from the
midpoint of its motion does its speed equal
one half of its maximum speed?
Answer in units of cm.
with an amplitude of 2.7 cm.
At what positive displacement from the
midpoint of its motion does its speed equal
one half of its maximum speed?
Answer in units of cm.
Answers
GPT-4o mini
In simple harmonic motion (SHM), the maximum speed \( v_{\text{max}} \) can be expressed as:
\[
v_{\text{max}} = A \omega
\]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.
Given that the amplitude \( A \) is 2.7 cm, the maximum speed is:
\[
v_{\text{max}} = 2.7 \, \text{cm} \cdot \omega
\]
The speed at a displacement \( x \) from the midpoint in SHM can be calculated using the formula:
\[
v = \sqrt{v_{\text{max}}^2 - \left( \frac{2\pi}{T} x \right)^2}
\]
It is also important to note that angular frequency \( \omega = \frac{2\pi}{T} \), but for the sake of this problem, we can just remain with \( \omega \).
Now, we need to find the displacement \( x \) where the speed \( v \) equals \( \frac{1}{2} v_{\text{max}} \):
\[
\frac{1}{2} v_{\text{max}} = \sqrt{v_{\text{max}}^2 - \left( \frac{2\pi}{T} x \right)^2}
\]
Substituting \( v_{\text{max}} = A \omega \):
\[
\frac{1}{2} (A \omega) = \sqrt{(A \omega)^2 - \left(\frac{2\pi}{T} x\right)^2}
\]
Squaring both sides:
\[
\left(\frac{1}{2} A \omega\right)^2 = (A \omega)^2 - \left(\frac{2\pi}{T} x\right)^2
\]
Expanding and rearranging gives:
\[
\frac{1}{4} A^2 \omega^2 = A^2 \omega^2 - \left(\frac{2\pi}{T} x\right)^2
\]
\[
\left(\frac{2\pi}{T} x\right)^2 = A^2 \omega^2 - \frac{1}{4} A^2 \omega^2
\]
\[
\left(\frac{2\pi}{T} x\right)^2 = \left(1 - \frac{1}{4}\right) A^2 \omega^2
\]
\[
\left(\frac{2\pi}{T} x\right)^2 = \frac{3}{4} A^2 \omega^2
\]
Taking the square root:
\[
\frac{2\pi}{T} x = \sqrt{\frac{3}{4}} A \omega
\]
\[
x = \frac{\sqrt{3}}{4} A = \frac{\sqrt{3}}{4} \cdot 2.7 \, \text{cm}
\]
Calculating this gives:
\[
x \approx \frac{1.732}{4} \cdot 2.7 \approx 0.173 \cdot 2.7 \approx 0.467 \, \text{cm}
\]
Thus, the positive displacement from the midpoint at which the speed equals one half of its maximum speed is approximately:
\[
\boxed{2.066 \, \text{cm}}
\]
\[
v_{\text{max}} = A \omega
\]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.
Given that the amplitude \( A \) is 2.7 cm, the maximum speed is:
\[
v_{\text{max}} = 2.7 \, \text{cm} \cdot \omega
\]
The speed at a displacement \( x \) from the midpoint in SHM can be calculated using the formula:
\[
v = \sqrt{v_{\text{max}}^2 - \left( \frac{2\pi}{T} x \right)^2}
\]
It is also important to note that angular frequency \( \omega = \frac{2\pi}{T} \), but for the sake of this problem, we can just remain with \( \omega \).
Now, we need to find the displacement \( x \) where the speed \( v \) equals \( \frac{1}{2} v_{\text{max}} \):
\[
\frac{1}{2} v_{\text{max}} = \sqrt{v_{\text{max}}^2 - \left( \frac{2\pi}{T} x \right)^2}
\]
Substituting \( v_{\text{max}} = A \omega \):
\[
\frac{1}{2} (A \omega) = \sqrt{(A \omega)^2 - \left(\frac{2\pi}{T} x\right)^2}
\]
Squaring both sides:
\[
\left(\frac{1}{2} A \omega\right)^2 = (A \omega)^2 - \left(\frac{2\pi}{T} x\right)^2
\]
Expanding and rearranging gives:
\[
\frac{1}{4} A^2 \omega^2 = A^2 \omega^2 - \left(\frac{2\pi}{T} x\right)^2
\]
\[
\left(\frac{2\pi}{T} x\right)^2 = A^2 \omega^2 - \frac{1}{4} A^2 \omega^2
\]
\[
\left(\frac{2\pi}{T} x\right)^2 = \left(1 - \frac{1}{4}\right) A^2 \omega^2
\]
\[
\left(\frac{2\pi}{T} x\right)^2 = \frac{3}{4} A^2 \omega^2
\]
Taking the square root:
\[
\frac{2\pi}{T} x = \sqrt{\frac{3}{4}} A \omega
\]
\[
x = \frac{\sqrt{3}}{4} A = \frac{\sqrt{3}}{4} \cdot 2.7 \, \text{cm}
\]
Calculating this gives:
\[
x \approx \frac{1.732}{4} \cdot 2.7 \approx 0.173 \cdot 2.7 \approx 0.467 \, \text{cm}
\]
Thus, the positive displacement from the midpoint at which the speed equals one half of its maximum speed is approximately:
\[
\boxed{2.066 \, \text{cm}}
\]