What is the velocity of an object with maximum velocity 10 m/s when displacement is 4 m, given the amplitude is 6 m?
Understand the Problem
The question is asking for the velocity of an object in harmonic motion given its maximum velocity, displacement, and amplitude. To solve it, we can use the relationship between maximum velocity, displacement, and amplitude in simple harmonic motion.
Answer
The velocity of the object is approximately $7.45 \, \text{m/s}$.
Answer for screen readers
The velocity of the object is given by:
$$ v = \frac{10\sqrt{5}}{3} \approx 7.45 , \text{m/s} $$
Steps to Solve
- Identify the given values
We have:
- Maximum velocity, $v_{max} = 10 , \text{m/s}$
- Displacement, $x = 4 , \text{m}$
- Amplitude, $A = 6 , \text{m}$
- Use the formula for velocity in harmonic motion
The formula to find velocity $v$ of an object in simple harmonic motion at a given displacement is:
$$ v = v_{max} \sqrt{1 - \left(\frac{x}{A}\right)^2} $$
- Substitute the values into the formula
Now substitute the values into the formula:
$$ v = 10 \sqrt{1 - \left(\frac{4}{6}\right)^2} $$
- Calculate the fraction and simplify
First, calculate $\left(\frac{4}{6}\right)^2$:
$$ \left(\frac{4}{6}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9} $$
Thus,
$$ 1 - \frac{4}{9} = \frac{5}{9} $$
- Final calculation of velocity
Now plug this back into the velocity formula:
$$ v = 10 \sqrt{\frac{5}{9}} $$
Which can be further simplified to:
$$ v = 10 \cdot \frac{\sqrt{5}}{3} $$
The velocity of the object is given by:
$$ v = \frac{10\sqrt{5}}{3} \approx 7.45 , \text{m/s} $$
More Information
This calculation shows how the velocity varies depending on the displacement within the bounds of maximum amplitude in simple harmonic motion. Understanding this relationship is crucial for analyzing oscillatory systems.
Tips
- Incorrectly squaring the fraction: Always ensure to square the fraction accurately when calculating $\left(\frac{x}{A}\right)^2$.
- Forgetting to take the square root: Make sure to take the square root of the quantity after finding the difference.
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