Answer:
The maximum velocity is 1.58 m/s.
Explanation:
A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.
Spring constant, K = 100 N/m
mass, m = 0.1 kg
Amplitude, A = 5 cm = 0.05 m
Let the angular frequency is w.
The maximum velocity is
Answer:
v=1.295
Explanation:
What we are given:
a=5÷(3s^(1/3)+s^(5/2)) m/s^2
Start by using equation a ds = v dv
This problem requires a numeric method of solving. Therefore, you can integrate v ds normally, but you must use a different method for a ds The problem should look like this:
<em>a=2</em>
<em>b=1</em>
<em>x=5÷(3s^(1/3)+s^(5/2)) </em><em>m/s^2</em>
<em>dx=dv</em>
Integrate the left side the standard method.
<em>a=v</em>
<em>b=0</em>
<em>dx=dv</em>
<em>Integrating</em>
=v^2/2
Use Simpson's rule for the right site.
<em>a=b</em>
<em>b=a</em>
<em>x=f(x)</em>
f(x)=b-a/6*(f(a)+4f(a+b/2)+f(b)
If properly applied. you should now have the following equation:
v^2/2=5[(1/6*(0.25+4(0.162)+(0.106)]
=0.8376
Solve for v.
v=1.295