# Simple Harmonic Motion-Mass

• physgrl
In summary: But the point of showing the equations was to demonstrate how to use F = ma and F = kx to solve the problem. And yes, I agree that the equations could have been written a bit more clearly to show that x represents the displacement from equilibrium. In summary, to find the acceleration of a 0.2 kg object undergoing simple harmonic motion with a spring constant of k=10 N/m at the instant when it is -0.05 m away from equilibrium, we can use the equations F=ma and F=-kx to calculate the acceleration. By setting the forces of gravity and the spring equal to each other, we can find the value of x at this point and use it to calculate the acceleration, which is
physgrl

## Homework Statement

A 0.2 kg object is suspended from a spring with a spring constant of k=10 N/m and is undergoing simple harmonic motion. What is its acceleration of the object at the instant when it is -0.05 m away from equilibrium?

A. 1000 m/s2
B. 40 m/s2
C. 0.1 m/s2
D. 2.5 m/s2

F=ma
F=-kx
F=mg

## The Attempt at a Solution

F=F(gravity)+F(spring)
ma=mg+(-kx)
0.20kg*a=(0.20kg*9.81m/s^2)+(-(10N/m*-0.05m))
a=12.31m/s^2

The answer is supposed to be D

Yeah, this is a tricky question if you don't already know the answer.

The equilibrium point will shift due to the gravity force, so when the object is -0.05m away from equilibrium, this will not be the value of x.

To calculate the new equilibrium point, just solve for a=0. And then you can get the value of x which is -0.05m from the equilibrium point.

BruceW said:
Yeah, this is a tricky question if you don't already know the answer.

The equilibrium point will shift due to the gravity force, so when the object is -0.05m away from equilibrium, this will not be the value of x.

Sure it will, if x is the displacement from equilibrium.

A straightforward way to approach this problem is to use write an expression for the position versus time (use a sine or cosine function). Find the corresponding acceleration expression (a touch of simple calculus). Take the position expression and rearrange for an expression for t in terms of x. Use it to find acceleration in terms of x.

gneill said:
Sure it will, if x is the displacement from equilibrium.
I was referring to her/his? equations, where x is not the displacement from equilibrium. Your post isn't very helpful.

gneill said:
A straightforward way to approach this problem is to use write an expression for the position versus time (use a sine or cosine function). Find the corresponding acceleration expression (a touch of simple calculus). Take the position expression and rearrange for an expression for t in terms of x. Use it to find acceleration in terms of x.
Yep, that's another way of doing it.

## 1. What is simple harmonic motion?

Simple harmonic motion is a type of oscillatory motion in which the restoring force is directly proportional to the displacement from equilibrium and always directed towards the equilibrium position.

## 2. What is the equation for simple harmonic motion?

The equation for simple harmonic motion can be written as x = A*cos(ωt + φ), where x is the displacement from equilibrium, A is the amplitude, ω is the angular frequency, and φ is the phase constant.

## 3. How does mass affect simple harmonic motion?

The mass of an object does not affect the frequency or period of simple harmonic motion. However, it does affect the amplitude of the motion, with larger masses resulting in smaller amplitudes.

## 4. What is the relationship between mass and spring constant in simple harmonic motion?

The mass and spring constant have an inverse relationship in simple harmonic motion. This means that as the mass increases, the spring constant must also increase in order to maintain the same frequency of oscillation.

## 5. How is energy conserved in simple harmonic motion?

In simple harmonic motion, the total mechanical energy (kinetic + potential) is conserved. This means that as the object oscillates, the energy is constantly transferred between kinetic and potential forms, but the total amount remains constant.

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