A 1.20 kg mass on a horizontal spring oscillates on a frictionless surface with a displacement as a function of time given by x(t) = 0.075cos(4.16t – 2.45). (Units are standard units.)
a.) Find the time for one complete vibration.
b.) Find the force constant of the spring.
c.) Find the maximum force on the mass.
d.) Find the maximum speed of the mass.
e.) Find the position, speed and acceleration of the mass at t = 1.00 s.
f.) Find the total energy of the oscillating spring.
a. when is 4.16 t = 2 pi ? Then t = T =2 pi/4.16
b. 2 pi f = sqrt (k/m)
so
2 pi/T = 4.16 = sqrt (k/1.2)
c. F = -k x
we know k now and max x = .075
d. dx/dt max = (4.16)(.075)
e. we have everything but acceleration
v = -.075(4.16)(sin( ) )
a = - (4.16)^2 x
f. we have v max
(1/2) m v^2 is total when v = v max
To answer these questions, we will need to use the formula for simple harmonic motion:
x(t) = A * cos(ωt + φ)
where:
x(t) is the displacement as a function of time,
A is the amplitude (maximum displacement),
ω is the angular frequency, and
φ is the phase angle.
Given the equation x(t) = 0.075cos(4.16t – 2.45), we can see that:
Amplitude (A) = 0.075
Angular Frequency (ω) = 4.16
Phase Angle (φ) = -2.45
Now let's answer each question step-by-step:
a.) Find the time for one complete vibration.
The time for one complete vibration is equal to the period (T). The period is defined as the time it takes for one complete cycle of oscillation. It can be calculated using the formula:
T = (2π) / ω
Plugging in the value of ω = 4.16, we can calculate the period:
T = (2π) / 4.16
T ≈ 1.511 s
Therefore, the time for one complete vibration is approximately 1.511 seconds.
b.) Find the force constant of the spring.
The force constant of the spring, denoted by k, is related to the angular frequency (ω) by the formula:
k = m * ω^2
Given that the mass (m) is 1.20 kg, we can calculate the force constant:
k = 1.20 * (4.16)^2
k ≈ 20.69 N/m
Therefore, the force constant of the spring is approximately 20.69 N/m.
c.) Find the maximum force on the mass.
The maximum force (Fmax) on the mass can be calculated using Hooke's Law:
Fmax = k * A
Plugging in the values of k = 20.69 N/m and A = 0.075, we can calculate the maximum force:
Fmax = 20.69 * 0.075
Fmax ≈ 1.552 N
Therefore, the maximum force on the mass is approximately 1.552 N.
d.) Find the maximum speed of the mass.
The maximum speed of the mass occurs when the displacement is maximum, which is equal to the amplitude (A). We can find the maximum speed (vmax) using the formula:
vmax = A * ω
Plugging in the values of A = 0.075 and ω = 4.16, we can calculate the maximum speed:
vmax = 0.075 * 4.16
vmax ≈ 0.312 m/s
Therefore, the maximum speed of the mass is approximately 0.312 m/s.
e.) Find the position, speed, and acceleration of the mass at t = 1.00 s.
To find the position (x), speed (v), and acceleration (a) of the mass at t = 1.00 s, we plug in t = 1.00 into the equation.
x(1.00) = 0.075cos(4.16 * 1.00 – 2.45)
x(1.00) ≈ 0.075cos(1.71)
x(1.00) ≈ 0.047 m
The position of the mass at t = 1.00 s is approximately 0.047 m.
To find the speed and acceleration at t = 1.00 s, we take the derivative of x(t) with respect to time (t):
v(t) = dx(t)/dt = -A * ω * sin(ωt + φ)
a(t) = dv(t)/dt = -A * ω^2 * cos(ωt + φ)
Plugging in the values of A = 0.075, ω = 4.16, and φ = -2.45, we can calculate the speed and acceleration:
v(1.00) = -0.075 * 4.16 * sin(4.16 * 1.00 – 2.45)
v(1.00) ≈ -0.136 m/s
The speed of the mass at t = 1.00 s is approximately -0.136 m/s (negative indicates the direction).
a(1.00) = -0.075 * (4.16)^2 * cos(4.16 * 1.00 – 2.45)
a(1.00) ≈ -2.170 m/s^2
The acceleration of the mass at t = 1.00 s is approximately -2.170 m/s^2.
f.) Find the total energy of the oscillating spring.
The total energy of the oscillating spring is the sum of the kinetic energy (KE) and potential energy (PE) at any given time. The equations for KE and PE are as follows:
KE = (1/2) * m * v^2
PE = (1/2) * k * x^2
Plugging in the values of m = 1.20 kg, v = vmax = 0.312 m/s (maximum speed), k = 20.69 N/m (force constant), and x = A = 0.075 (maximum displacement), we can calculate the total energy:
KE = (1/2) * 1.20 * (0.312)^2
KE ≈ 0.058 J
PE = (1/2) * 20.69 * (0.075)^2
PE ≈ 0.058 J
The total energy of the oscillating spring is approximately 0.058 J.
a.) To find the time for one complete vibration, we need to determine the period of the oscillation. The period is the time it takes for the system to complete one full cycle of motion.
The period (T) can be found using the formula T = 2π/ω, where ω is the angular frequency. The angular frequency is determined by the coefficient of t in the equation x(t) = Acos(ωt + φ), where A is the amplitude and φ is the phase constant.
In this case, the coefficient of t is 4.16. So, ω = 4.16. Plugging this value into the formula, we have T = 2π/4.16. Evaluating this expression, we find the time for one complete vibration to be approximately 1.51 seconds.
b.) The force constant (k) of the spring can be found using the formula k = mω^2, where m is the mass and ω is the angular frequency.
In this case, the mass (m) is given as 1.20 kg and the angular frequency (ω) is 4.16. Plugging these values into the formula, we have k = 1.20 * (4.16)^2. Evaluating this expression, we find the force constant of the spring to be approximately 21.11 N/m.
c.) The maximum force on the mass can be found using the formula F_max = kA, where A is the amplitude of the oscillation.
In this case, the amplitude (A) is given as 0.075. Plugging this value along with the force constant (k = 21.11 N/m) into the formula, we have F_max = 21.11 * 0.075. Evaluating this expression, we find the maximum force on the mass to be approximately 1.58 N.
d.) The maximum speed of the mass can be found using the formula v_max = Aω, where A is the amplitude and ω is the angular frequency.
In this case, the amplitude (A) is given as 0.075 and the angular frequency (ω) is 4.16. Plugging these values into the formula, we have v_max = 0.075 * 4.16. Evaluating this expression, we find the maximum speed of the mass to be approximately 0.312 m/s.
e.) To find the position, speed, and acceleration of the mass at t = 1.00 s, we can evaluate the given equation x(t) = 0.075cos(4.16t – 2.45) at t = 1.00 s.
Plugging t = 1.00 s into the equation, we have x(1.00) = 0.075cos(4.16 * 1.00 – 2.45). Evaluating this expression, we find the position of the mass at t = 1.00 s to be approximately 0.0375 m (or 3.75 cm).
To find the speed, we can take the derivative of x(t) with respect to time (t). The derivative of cos(4.16t – 2.45) is -4.16sin(4.16t – 2.45). Plugging t = 1.00 s into this expression, we find the speed at t = 1.00 s to be approximately -0.344 m/s.
To find the acceleration, we can take the second derivative of x(t) with respect to time (t). The second derivative of cos(4.16t – 2.45) is -4.16^2cos(4.16t – 2.45). Plugging t = 1.00 s into this expression, we find the acceleration at t = 1.00 s to be approximately -18.21 m/s^2.
f.) To find the total energy of the oscillating spring, we can use the formula E_total = (1/2)kA^2, where k is the force constant and A is the amplitude.
In this case, the force constant (k) is given as 21.11 N/m and the amplitude (A) is given as 0.075. Plugging these values into the formula, we have E_total = (1/2) * 21.11 * (0.075)^2. Evaluating this expression, we find the total energy of the oscillating spring to be approximately 0.059 J.