Solve Simple Harmonic Motion Homework for Mass m on Spring s

[Lavace]

Homework Statement

A spring with stiffness s is suspended vertically with a mass m attached to it's free end. When the mass comes to rest the spring is found to have been stretched by 0.0981m. The mass is then pulled down a further 0.1m, released from rest and found to execute SHM.

We're given the general form of the SHM equation, i.e x(t) = Acos(wt + psi).

i) Calculate numerical values for the following: The amplitute, constant phase angle, frequency, period, maximum and minimum speed of the mass, the maximum kinetic energy for a mass of 1kg.

Homework Equations

F =- kx
U = 1/2*kx^2

The Attempt at a Solution

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I don't particularly know where to start, or rather where I intended to start is incorrect.

I was thinking that we'd start off by calculating the spring constant (stiffness), k (called s in the script). But where this fits into finding the amplitude I'm not sure.
Is the amplitude just 0.1m (as that is where the mass was released).

What is meant by constant phase angle? Psi?

To find the period, we will have to know omega(0), which is equal to 2PI/T, where T is the period. Omega(0) is equal to k/m, but we don't know the mass, we could substitute in Newton II (F=ma), but I'm not sure where this will lead.

Any pushes in the correct direction is a massive thank you!

 

[kuruman]

Can you find an expression for the distance by which a spring of stiffness s is stretched when mass m is attached to it?

 

[Lavace]

So here's my new solution:

F = mg - kd - kx , where d is the extension the string was stretched by.

Because mg = kd (from the equilibrium balance), F = -kx

We can then say ma = -kx, and hence a = -(k/m)x, where w = SQRT(k/m)

But then, why were we given the extension of when the mass was placed on the spring? Was this to calculate the stiffness for the latter questions?

 

[kuruman]

How will you find the amplitude?

 

[paranormal]

Hello, thank you for your question! I would approach this problem by first defining the variables and equations that we will need to use.

Variable definitions:
- m = mass of the object (given)
- s = spring stiffness (given)
- x(t) = position of the mass as a function of time (SHM equation)
- A = amplitude (unknown)
- w = angular frequency (unknown)
- t = time (unknown)
- psi = constant phase angle (unknown)

Equations:
- SHM equation: x(t) = Acos(wt + psi)
- Newton's Second Law: F = ma
- Hooke's Law: F = -kx
- Potential energy of a spring: U = 1/2 * k * x^2
- Kinetic energy: KE = 1/2 * m * v^2 (where v is the velocity of the mass)

Now, let's start by finding the amplitude. As you correctly mentioned, the amplitude is the maximum displacement of the mass from its equilibrium position. In this case, the mass was pulled down by 0.1m from its equilibrium position, so the amplitude is also 0.1m.

Next, we can use Newton's Second Law and Hooke's Law to find the angular frequency (w). From Newton's Second Law, we know that F = ma, where F is the force acting on the mass, m is the mass, and a is the acceleration. In this case, the only force acting on the mass is the force from the spring, which is given by Hooke's Law: F = -kx. So we can set these two equations equal to each other and solve for a: ma = -kx. We can then substitute in the given values for m and x (0.1m), and solve for k: k = ma/x = (m * 9.8 m/s^2)/0.1m = 98 m/s^2.

Now that we have the value for k, we can use the equation w = sqrt(k/m) to find the angular frequency. Substituting in the values for k and m, we get w = sqrt(98 m/s^2 / m) = sqrt(98/s) = 9.9 rad/s.

To find the constant phase angle (psi), we can use the information given in the problem.