Intro to Simple Harmonic Motion (Horizontal Springs) Video Lessons

Video Thumbnail

Concept

Problem: Learning Goal: To understand the application of the general harmonic equation to finding the acceleration of a spring oscillator as a function of time.One end of a spring with spring constant k is attached to the wall. The other end is attached to a block of mass m. The block rests on a frictionless horizontal surface. The equilibrium position of the left side of the block is defined to be x = 0. The length of the relaxed spring is L.(Figure 1)The block is slowly pulled from its equilibrium position to some position xinit > 0 along the x axis. At time t = 0, the block is released with zero initial velocity.The goal of this problem is to determine the acceleration of the block a(t) as a function of time in terms of k, m, and xinit.It is known that a general solution for the position of a harmonic oscillator isx(t) = C cos(ωt) + S sin(ωt),where C, S, and ω are constants. (Figure 2)Your task, therefore, is to determine the values of C, S, and ω in terms of k, m, and xinit and then use the connection between x(t) and a(t) to find the acceleration.QUESTION: Using the fact that acceleration is the second derivative of position, find the acceleration of the block a(t) as a function of time. Express your answer in terms of ω, t, and x(t).

FREE Expert Solution

Derivative of sine/cosine:

ddtsin(at)=acos(at)ddtcos(at)=-asin(at)

We'll determine the derivatives then substitute initial conditions to solve for C and S.

x(t) = C cos(ωt) + S sin(ωt)

v(t) = (d/dt)x(t) = (d/dt)(C cos(ωt) + S sin(ωt)) = -ωC sin(ωt) + ωS cos(ωt)

85% (220 ratings)
View Complete Written Solution
Problem Details

Learning Goal: To understand the application of the general harmonic equation to finding the acceleration of a spring oscillator as a function of time.

One end of a spring with spring constant k is attached to the wall. The other end is attached to a block of mass m. The block rests on a frictionless horizontal surface. The equilibrium position of the left side of the block is defined to be x = 0. The length of the relaxed spring is L.(Figure 1)

The block is slowly pulled from its equilibrium position to some position xinit > 0 along the x axis. At time t = 0, the block is released with zero initial velocity.

The goal of this problem is to determine the acceleration of the block a(t) as a function of time in terms of k, m, and xinit.

It is known that a general solution for the position of a harmonic oscillator is

x(t) = C cos(ωt) + S sin(ωt),

where C, S, and ω are constants. (Figure 2)

Your task, therefore, is to determine the values of C, S, and ω in terms of k, m, and xinit and then use the connection between x(t) and a(t) to find the acceleration.



QUESTION: Using the fact that acceleration is the second derivative of position, find the acceleration of the block a(t) as a function of time. Express your answer in terms of ω, t, and x(t).

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Intro to Simple Harmonic Motion (Horizontal Springs) concept. You can view video lessons to learn Intro to Simple Harmonic Motion (Horizontal Springs). Or if you need more Intro to Simple Harmonic Motion (Horizontal Springs) practice, you can also practice Intro to Simple Harmonic Motion (Horizontal Springs) practice problems.