A spring id vertically, as shown below, and a 500g mass is hung from its lower end. The EQUILIBRIUM extension of the spring is observed to be 10.0cm.
(1) What is the spring constant of the spring?
ANSWER: 49N/m
(B) If the is pulled down by 5.00cm and released, what is the frequency of the subsequent oscillations. ANSWER: 1.58
(C) What ist he velocity of the mass at time t=2.5 sec?
(D) What is the acceleration of the mass at time t=4.75 sec?
(E) What is the net force acting on the mass at time t=5.0sec?
(F) What are the potential and kinetic energies of the mass when the displacement 7 =-1.5cm?
I NEED HELP ON C THROUGH F. THANK YOU
C) To find the velocity of the mass at time t = 2.5 seconds, we need to use the equations of motion for simple harmonic motion. The equation relating velocity and displacement in simple harmonic motion is given by:
v = ω √(A^2 - x^2)
where v is the velocity, ω is the angular frequency, A is the amplitude, and x is the displacement.
In this case, the amplitude A is the equilibrium extension of the spring, which is 10.0 cm or 0.1 m. We need to convert this to SI units.
A = 0.1 m
The angular frequency ω is related to the spring constant k by the equation:
ω = √(k/m)
where m is the mass in kilograms.
In this case, m is 500 g or 0.5 kg.
m = 0.5 kg
To find the spring constant k, we can use Hooke's Law:
F = kx
where F is the force, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the force F is the weight of the mass:
F = mg
F = 0.5 kg × 9.8 m/s^2
Now, we can solve for k:
k = F/x
k = (0.5 kg × 9.8 m/s^2) / 0.1 m
Once we have k, we can find the angular frequency ω:
ω = √(k/m)
Now, we can use the equation for velocity to find the velocity at time t = 2.5 seconds:
v = ω √(A^2 - x^2)
Plug in the values and calculate the velocity.
D) To find the acceleration of the mass at time t = 4.75 seconds, we can use the equation for acceleration in simple harmonic motion:
a = -ω^2x
where a is the acceleration, ω is the angular frequency, and x is the displacement from the equilibrium position.
We already know the value of ω from part C.
Once we have ω, we can use the equation for acceleration to find the acceleration at time t = 4.75 seconds.
E) To find the net force acting on the mass at time t = 5.0 seconds, we can use Hooke's Law.
We can use the equation for force in simple harmonic motion:
F = -kx
where F is the force, k is the spring constant, and x is the displacement from the equilibrium position.
We already know the value of k from part C.
Once we have k, we can use the equation for force to find the net force at time t = 5.0 seconds.
F) To find the potential and kinetic energies of the mass when the displacement x = -1.5 cm or -0.015 m, we can use the formulas for potential and kinetic energies in simple harmonic motion.
The potential energy (PE) is given by:
PE = 0.5kx^2
where k is the spring constant and x is the displacement from the equilibrium position.
We already know the value of k from part C.
Once we have k, we can use the equation for potential energy to find the potential energy when x = -0.015 m.
The kinetic energy (KE) is given by:
KE = 0.5mv^2
where m is the mass and v is the velocity.
We need to use the equation for velocity from part C to find the velocity when x = -0.015 m.
Once we have m and v, we can use the equation for kinetic energy to find the kinetic energy when x = -0.015 m.