The displacement of a block of mass 439g
439
g
attached to a spring whose spring constant is 66N/m
66
N
m
is given by x=Asin(ωt)
x
A
s
i
n
ω
t
where A=12cm
A
12
c
m
. In the first complete cycle find the values of x
x
and t
t
at which the kinetic energy is equal to one half the potential energy.
First position: cm...... First time:
To find the values of x and t at which the kinetic energy is equal to one half the potential energy in the first complete cycle, we need to understand the equations for kinetic energy (KE) and potential energy (PE) of the block attached to the spring.
The equation for kinetic energy is given by KE = (1/2)mv^2, where m is the mass of the block and v is its velocity.
The equation for potential energy of a spring is given by PE = (1/2)kx^2, where k is the spring constant and x is the displacement of the block from its equilibrium position.
In the given displacement equation x = Asin(ωt), A is the amplitude of the oscillation, ω is the angular frequency, and t is time.
Since we want to find the position and time at which KE is equal to (1/2)PE, we can set up the equation:
(1/2)mv^2 = (1/2)kx^2
Given:
m = 439g = 0.439kg (convert grams to kilograms)
k = 66N/m
A = 12cm = 0.12m (convert centimeters to meters)
First, let's find ω:
ω = √(k/m)
ω = √(66N/m / 0.439kg)
ω ≈ 7.82 rad/s
Now, let's find the first position and time where KE is equal to (1/2)PE.
To find the position at which KE is equal to (1/2)PE, we need to substitute x into the potential energy equation and solve for x when KE = (1/2)PE.
(1/2)mv^2 = (1/2)kx^2
0.5 * 0.439kg * v^2 = 0.5 * 66N/m * x^2
v^2 = (66N/m * x^2) / 0.439kg
Since v = ωAcos(ωt), substitute v into the equation:
(7.82 rad/s)^2 = (66N/m * x^2) / 0.439kg
Now, substitute x = Asin(ωt) into the equation:
(7.82 rad/s)^2 = (66N/m * (0.12m * sin(7.82t))^2) / 0.439kg
Simplify the equation and solve for t.
Once we find the value of t, we can substitute it back into x = Asin(ωt) to find the value of x.