A body is suspended vertically from an ideal spring of spring constant 3.4 N/m. The spring is initially in its relaxed position. The body is then released and oscillates about its equilibrium position. The motion is described by the following equation.
Without providing the equation, I am unable to explain the specific motion described. However, I can explain how to analyze the motion of a body suspended from an ideal spring.
To describe the motion of the system, we need to understand Hooke's Law and the equation of motion for simple harmonic motion (SHM).
1. Hooke's Law:
Hooke's Law states that the force exerted by an ideal spring is directly proportional to the displacement from its equilibrium position. Mathematically, it can be expressed as:
F = -kx
where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
2. Equation of motion for SHM:
The equation of motion for SHM can be expressed as:
a = -ω^2x
where a is the acceleration of the body, ω is the angular frequency, and x is the displacement from the equilibrium position. The negative sign represents the direction opposite to the displacement.
By combining Hooke's Law and the equation of motion for SHM, we can obtain the differential equation that governs the motion of the system:
m(d^2x/dt^2) = -kx
where m is the mass of the body, and (d^2x/dt^2) represents the second derivative of x with respect to time.
Solving this differential equation will give us the equation that describes the motion of the body.
In your specific case, you mentioned that the spring constant is 3.4 N/m. To determine the equation describing the motion, you need to provide additional information such as the mass of the body or any initial conditions (e.g., initial displacement or velocity). With this information, you can solve the differential equation and obtain the equation describing the motion.
To describe the motion of the body oscillating about its equilibrium position, we can use the equation for simple harmonic motion:
x(t) = A * cos(ωt + φ)
Where:
- x(t) is the displacement from the equilibrium position at time t.
- A is the amplitude of the oscillation.
- ω is the angular frequency of the oscillation.
- φ is the phase constant or initial phase.
In this case, since the body is initially at rest when released, we can assume that the displacement at t = 0 is maximum, i.e. A = maximum displacement. The equation becomes:
x(t) = A * cos(ωt)
To find the angular frequency (ω), we can use the formula:
ω = √(k / m)
Where:
- k is the spring constant (3.4 N/m in this case).
- m is the mass of the body.
Since the body is suspended vertically, we can assume that the weight of the body is equal to the force exerted by the spring at equilibrium. The weight of the body can be calculated using the formula:
Weight = m * g
Where:
- m is the mass.
- g is the acceleration due to gravity (9.8 m/s²).
Since the weight is equal to the force exerted by the spring at equilibrium, we have:
k * A = m * g
Substituting k = 3.4 N/m and g = 9.8 m/s², we can solve for A:
3.4 * A = m * 9.8
A = (m * 9.8) / 3.4
So A is equal to (m * 9.8) / 3.4.
Now that we have A and ω, we can write the equation for the motion of the body as:
x(t) = ((m * 9.8) / 3.4) * cos(ωt)
Note: The equation assumes no damping or external forces acting on the body. It also assumes small amplitude oscillations, where the restoring force is linearly proportional to the displacement.