A 3.5 kg object is suspended from a spring of spring constant 725 N/m. Starting from the equilibrium position, the object has an initial speed of 7.9 m/s. What is the maximum displacement of the object during its subsequent motion? (In cm)
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To find the maximum displacement of the object during its subsequent motion, we need to use the principles of simple harmonic motion.
First, let's find the potential energy stored in the spring when the object is at maximum displacement. The potential energy stored in a spring is given by the formula:
Potential Energy (PE) = (1/2) * k * x^2
where k is the spring constant and x is the displacement from the equilibrium position. In this case, we want to find when the potential energy is at its maximum, so we need to maximize x.
At the maximum displacement, all the initial kinetic energy will be converted into potential energy. So we can write the equation for the initial kinetic energy as:
Kinetic Energy (KE) = (1/2) * m * (v^2)
where m is the mass of the object and v is the initial velocity.
Setting the initial kinetic energy equal to the potential energy at maximum displacement, we have:
(1/2) * m * (v^2) = (1/2) * k * x^2
Simplifying, we get:
m * (v^2) = k * x^2
Now, we can plug in the given values:
m = 3.5 kg
v = 7.9 m/s
k = 725 N/m
Plugging in these values, we get:
3.5 * (7.9^2) = 725 * x^2
Solving for x^2, we have:
x^2 = (3.5 * (7.9^2)) / 725
x^2 = 48.090344827586206
Taking the square root of both sides, we get:
x = √(48.090344827586206)
x ≈ 6.93 m
Finally, we convert the displacement to centimeters by multiplying by 100:
Maximum displacement = 6.93 * 100 cm
Maximum displacement ≈ 693 cm
Therefore, the maximum displacement of the object during its subsequent motion is approximately 693 cm.