A 0.50 kg mass at the end of a spring vibrates 6.0 times per second with an amplitude of 0.18 m.
(a) Determine the velocity when it passes the equilibrium point.
1Your answer is incorrect. m/s
(b) Determine the velocity when it is 0.02 m from equilibrium.
2 m/s
(c) Determine the total energy of the system.
3 J
(d) Complete the equation describing the motion of the mass, assuming that x was a maximum at t = 0.
4 m cos[ ( 5 rad/s )t ]
(a) To determine the velocity when the mass passes the equilibrium point, we can use the equation for the velocity of an object undergoing simple harmonic motion:
v = ωA sin(ωt)
Where:
v = velocity
ω = angular frequency = 2πf
A = amplitude
t = time
In this case, the angular frequency (ω) can be calculated as:
ω = 2πf = 2π(6.0 Hz) = 12π rad/s
The amplitude (A) is given as 0.18 m.
When the mass passes the equilibrium point, the displacement (x) is zero. Therefore, the equation becomes:
v = ωA sin(ωt) = (12π rad/s)(0.18 m)sin(ωt)
To find the velocity when it passes the equilibrium point, we substitute t = 0 into the equation:
v = (12π rad/s)(0.18 m)sin(0) = 0 m/s
So the velocity when it passes the equilibrium point is 0 m/s.
(b) To determine the velocity when the mass is 0.02 m from equilibrium, we need to find the time at which it reaches that displacement. Using the equation for displacement in simple harmonic motion:
x = A cos(ωt)
Where:
x = displacement
A = amplitude
ω = angular frequency
t = time
Rearranging the equation to solve for t:
t = (1/ω)arccos(x/A)
Plugging in the given values:
t = (1/(12π rad/s))arccos(0.02 m/0.18 m) ≈ 0.006535 s
Now, we can find the velocity when the mass is 0.02 m from equilibrium using the equation for velocity:
v = ωA sin(ωt) = (12π rad/s)(0.18 m)sin(12π rad/s)(0.006535 s)
Calculating this expression:
v ≈ 0.162 m/s
So the velocity when it is 0.02 m from equilibrium is approximately 0.162 m/s.
(c) The total energy of the system can be calculated as the sum of the kinetic energy (KE) and the potential energy (PE). In simple harmonic motion, the total energy remains constant.
The kinetic energy is given by:
KE = (1/2)mv^2
Where:
m = mass = 0.50 kg
v = velocity
Using the velocity calculated in part (b):
KE = (1/2)(0.50 kg)(0.162 m/s)^2
Simplifying:
KE ≈ 0.0104 J
The potential energy is given by:
PE = (1/2)kx^2
Where:
k = spring constant
x = displacement
The spring constant (k) is not given in the question, so we cannot determine the exact value at this time.
(d) The equation describing the motion of the mass is:
x(t) = A cos(ωt)
Where:
A = amplitude
ω = angular frequency
t = time
Substituting the given values into the equation:
x(t) = 0.18 m cos((5 rad/s)t)
So the equation describing the motion of the mass, assuming that x was a maximum at t = 0, is:
x(t) = 0.18 m cos((5 rad/s)t)
To answer these questions, we can use the equations of motion for a simple harmonic oscillator. These equations describe the motion of a mass-spring system.
The equation for the velocity of the mass at any given time is given by:
v(t) = Aω cos(ωt + φ)
Where:
- v(t) is the velocity of the mass at time t
- A is the amplitude of the motion (0.18 m in this case)
- ω is the angular frequency of the motion (2πf, where f is the frequency in Hz)
- φ is the phase angle, which determines the starting position of the motion (0 in this case, as x was a maximum at t = 0)
(a) To find the velocity when the mass passes the equilibrium point, we need to find the time at which the displacement is zero (equilibrium point), and then calculate the velocity at that time.
The time period (T) is the reciprocal of the frequency:
T = 1/f = 1/6.0 Hz = 0.167 s
At the equilibrium point, the displacement (x) is zero.
Using the equation x(t) = A cos(ωt + φ), we can solve for t:
0 = 0.18 cos(2π(6.0)t + 0)
Simplifying the equation, we get:
cos(2π(6.0)t) = 0
To find the time when the cosine function is equal to zero, we need to find the values of t that satisfy:
2π(6.0)t = π/2 + nπ, where n is an integer
Solving for t, we get:
t = (1/2π)(π/2 + nπ) / (6.0)
Plugging in n = 0 (for the first time the motion passes through the equilibrium point), we get:
t = (1/2π)(π/2) / (6.0) = 0.0262 s
Now, we can calculate the velocity at this time using the equation for velocity:
v(t) = Aω cos(ωt + φ)
v(0.0262 s) = 0.18 * 2π * 6.0 cos(2π(6.0)(0.0262) + 0)
v(0.0262 s) ≈ 2.58 m/s
Therefore, the velocity when the mass passes the equilibrium point is approximately 2.58 m/s.
(b) To find the velocity when the mass is 0.02 m from equilibrium, we need to find the time at which the displacement is 0.02 m, and then calculate the velocity at that time.
Using the equation x(t) = A cos(ωt + φ), we can solve for t:
0.02 = 0.18 cos(2π(6.0)t + 0)
Simplifying the equation, we get:
cos(2π(6.0)t) = 0.1111
To find the time when the cosine function is equal to 0.1111, we need to find the values of t that satisfy:
2π(6.0)t = arccos(0.1111) + 2nπ, where n is an integer
Solving for t, we get:
t = (1/2π)(arccos(0.1111) + 2nπ ) / (6.0)
Plugging in n = 0 (for the first time the displacement is 0.02 m), we get:
t = (1/2π)(arccos(0.1111)) / (6.0) ≈ 0.0542 s
Now, we can calculate the velocity at this time using the equation for velocity:
v(t) = Aω cos(ωt + φ)
v(0.0542 s) = 0.18 * 2π * 6.0 cos(2π(6.0)(0.0542) + 0)
v(0.0542 s) ≈ 2.0 m/s
Therefore, the velocity when the mass is 0.02 m from equilibrium is approximately 2.0 m/s.
(c) The total energy of the system can be calculated using the equation:
E = (1/2)kA^2
Where:
- E is the total energy of the system
- k is the spring constant
- A is the amplitude of the motion
To calculate E, we need the value of the spring constant (k). The spring constant can be determined using the equation:
ω = √(k/m)
Where:
- ω is the angular frequency of the motion
- k is the spring constant
- m is the mass
Using the given values, we can solve for k:
(2πf)^2 = k/m
(2π(6.0))^2 = k/0.50 kg
k = (2π(6.0))^2 * 0.50 kg = 226.195 N/m
Now, we can calculate the total energy:
E = (1/2)(226.195 N/m)(0.18 m)^2
E ≈ 3.25 J
Therefore, the total energy of the system is approximately 3.25 J.
(d) The equation describing the motion of the mass, assuming that x was a maximum at t = 0, is given by:
x(t) = A cos(ωt + φ)
Plugging in the values we have:
x(t) = 0.18 cos(2π(6.0)t + 0)
Therefore, the equation describing the motion of the mass is:
x(t) = 0.18 cos(12πt)