A 250 g object attached to a spring oscillates on a frictionless horizontal table with the frequency of 4 Hz and an amplitude of 20 cm. Calculate
a) maximum potential energy of the system
b) the displacement of the object when the potential energy is one-half of the maximum
c) the potential energy when the displacement is 10 cm.
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To calculate the maximum potential energy of the system, we can use the formula for potential energy in a spring:
a) Maximum potential energy (PE_max) = (1/2) * k * A^2
Where:
PE_max is the maximum potential energy
k is the spring constant
A is the amplitude of the oscillation
First, let's find the spring constant (k) using the formula for the frequency of the oscillation in terms of the spring constant and the mass:
f = (1 / 2π) * sqrt(k / m)
Rearranging the formula, we can solve for k:
k = (4π^2 * m * f^2)
Given:
m = 250 g = 0.25 kg (convert grams to kilograms)
f = 4 Hz
k = (4 * π^2 * 0.25 * 4^2)
= 16 * π^2
Now we can calculate the maximum potential energy:
PE_max = (1/2) * (16π^2) * (0.20)^2
= 1.6 * π^2 J (Joules)
b) To find the displacement (x) when the potential energy is one-half of the maximum, we can use the formula for potential energy and rearrange it to solve for x:
PE = (1/2) * k * x^2
Rearranging the formula, we get:
x = sqrt((2 * PE) / k)
Given:
PE = (1/2) * PE_max
x = sqrt((2 * (1/2) * (1.6 * π^2)) / (16 * π^2))
x = sqrt(0.1)
= 0.3162 m (meters)
c) To find the potential energy when the displacement is 10 cm, we can use the formula for potential energy:
PE = (1/2) * k * x^2
Given:
x = 10 cm = 0.1 m (convert centimeters to meters)
PE = (1/2) * (16π^2) * (0.1)^2
= 0.08 * π^2 J (Joules)
Therefore:
a) The maximum potential energy of the system is 1.6 * π^2 J (Joules).
b) The displacement of the object when the potential energy is one-half of the maximum is 0.3162 m (meters).
c) The potential energy when the displacement is 10 cm is 0.08 * π^2 J (Joules).