# Displacement in Simple Harmonic Motion (1 Viewer)

### Users Who Are Viewing This Thread (Users: 0, Guests: 1)

#### chukie

1. Hi! I am new to this forum. I was wondering if someone could help me with this problem. Thanks in advance!

A 0.27 kg mass is suspended on a spring that stretches a distance of 4.9 cm. The mass is then pulled down an additional distance of 12.5 cm and released. What's the displacement from the equilibrium position with the mass attached (in cm) after 0.42 s? Take up to be positive and use g = 9.81 m/s2.

2. Relevant equations
F=kx
ω(angular velocity)= (k/x)^0.5

3. The attempt at a solution

First I solved for k:
k=mg/x
k=(0.27*9.81)/4.9
k=0.54

Then I solved for the angular velocity:
ω=(0.54/0.27)^0.5
ω=squareroot 2

Then I tried to come up with an equation for displacement using the format:
x(t) = Asin(ωt)

i know that ω=root2
t=0.42

but how do i solve for A?? or what is A? should it be 4.9 or 12.5 or the sum of the two btw please correct me if im wrong in my calculation. thanks!

#### Snazzy

4.9 cm is the location of the equilibrium position. They're asking you to find the displacement after t seconds relative to the equilibrium position, so set your equilibrium position to 0. From the equilibrium position, they extended the displacement by 12.5 cm.

#### physixguru

You know anything about S.H.M equations?

#### chukie

thanks! i got the answer now =)

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving