# Simple Harmonic Motion displacement

• VincentweZu
In summary, the question asks at what displacement from equilibrium is the speed half of the maximum value in a spring system undergoing simple harmonic motion with amplitude A. The solution provided in the text is incorrect as it equates an amplitude of half the current amplitude to half the maximum velocity, instead of finding the actual displacement at which the velocity is half the maximum value. The correct approach involves using the law of conservation of energy and solving for the displacement using the equation \frac{1}{2}k(Δx)2= \frac{1}{2}mv2. Therefore, the correct answer is (\sqrt{3}/2)A.
VincentweZu

## Homework Statement

A mass m is attached to a horizontal spring of spring constant k. The spring oscillates in simple harmonic motion with amplitude A. Answer the following in terms of A.

At what displacement from equilibrium is the speed half of the maximum value?

## Homework Equations

PE1 + KE1 = PE2 + KE2

## The Attempt at a Solution

$\frac{1}{2}$kA2 = $\frac{1}{2}$mvmax2
kA2 = mvmax2
kA2/m = vmax2

$\frac{1}{2}$kA2 = $\frac{1}{2}$mv2 + $\frac{1}{2}$kΔx2 let v = vmax/2

kA2 = m(vmax/2)2 + kΔx2
kA2 = m(vmax2)/4 + kΔx2
kA2 = m(kA2/m)/4 + kΔx2
kA2 = kA2/4 + kΔx2
A2 = A2/4 + Δx2
(3/4)A2 = Δx2
Δx = ($\sqrt{3}$/2)A

the answer that I arrived at was ($\sqrt{3}$/2)A, however the answer is 1/2. If I subbed in arbitrary numbers for A, m, and k and let Δx = ($\sqrt{3}$/2)A then the velocity that results is half the max velocity.

I would like someone to confirm my answer or find something wrong with it so that I know which is the right answer. Thanks.

VincentweZu said:

## Homework Statement

A mass m is attached to a horizontal spring of spring constant k. The spring oscillates in simple harmonic motion with amplitude A. Answer the following in terms of A.

At what displacement from equilibrium is the speed half of the maximum value?

## Homework Equations

PE1 + KE1 = PE2 + KE2

## The Attempt at a Solution

$\frac{1}{2}$kA2 = $\frac{1}{2}$mvmax2
kA2 = mvmax2
kA2/m = vmax2

$\frac{1}{2}$kA2 = $\frac{1}{2}$mv2 + $\frac{1}{2}$kΔx2 let v = vmax/2

kA2 = m(vmax/2)2 + kΔx2
kA2 = m(vmax2)/4 + kΔx2
kA2 = m(kA2/m)/4 + kΔx2
kA2 = kA2/4 + kΔx2
A2 = A2/4 + Δx2
(3/4)A2 = Δx2
Δx = ($\sqrt{3}$/2)A

the answer that I arrived at was ($\sqrt{3}$/2)A, however the answer is 1/2. If I subbed in arbitrary numbers for A, m, and k and let Δx = ($\sqrt{3}$/2)A then the velocity that results is half the max velocity.

I would like someone to confirm my answer or find something wrong with it so that I know which is the right answer. Thanks.

I think the difficulty arises when you say $\frac{1}{2}$kA2 = $\frac{1}{2}$mvmax2

certainly those two quantities are equal in size, but at no time during the oscillation do both conditions occur at the same time, so a calculation based on them may not work.

Definitely there is no instant during oscillation where both these conditions occur, however, the equation which I have used - the conservation of energy, does not require that they do occur at the same time. In fact, all the law of conservation of energy states is that the total energy in one state is the same as the total energy in another so long as the system is a closed system.

In any case, the solution to this answer provided in the text doesn't convince me that at a displacement of (1/2)A would yield a velocity which is half the maximum velocity.

Basically the solution in the text involved only one equation which was:

$\frac{1}{2}$kA$^{2}$ = $\frac{1}{2}$mvmax2

They solve for vmax giving

vmax = A$\sqrt{\frac{k}{m}}$

vmax $\alpha$ A

Being directly proportional (1/2)A would lead to (1/2)vmax

I am not convinced by this solution as instead of finding at which state in the oscillation the velocity would be half its maximum, the solution says that an amplitude of half the current amplitude would half the maximum velocity.

VincentweZu said:
I am not convinced by this solution as instead of finding at which state in the oscillation the velocity would be half its maximum, the solution says that an amplitude of half the current amplitude would half the maximum velocity.

You're right. It looks like they either misinterpreted their own question or failed to clearly state their intention.

VincentweZu said:

## Homework Statement

A mass m is attached to a horizontal spring of spring constant k. The spring oscillates in simple harmonic motion with amplitude A. Answer the following in terms of A.

At what displacement from equilibrium is the speed half of the maximum value?

## Homework Equations

PE1 + KE1 = PE2 + KE2

## The Attempt at a Solution

$\frac{1}{2}$kA2 = $\frac{1}{2}$mvmax2
kA2 = mvmax2
kA2/m = vmax2

$\frac{1}{2}$kA2 = $\frac{1}{2}$mv2 + $\frac{1}{2}$kΔx2 let v = vmax/2

kA2 = m(vmax/2)2 + kΔx2
kA2 = m(vmax2)/4 + kΔx2
kA2 = m(kA2/m)/4 + kΔx2
kA2 = kA2/4 + kΔx2
A2 = A2/4 + Δx2
(3/4)A2 = Δx2
Δx = ($\sqrt{3}$/2)A

the answer that I arrived at was ($\sqrt{3}$/2)A, however the answer is 1/2. If I subbed in arbitrary numbers for A, m, and k and let Δx = ($\sqrt{3}$/2)A then the velocity that results is half the max velocity.

I would like someone to confirm my answer or find something wrong with it so that I know which is the right answer. Thanks.

ur equation 1/2k A^2=1/2mv^2+... is wrong.
it should be 1/2k(Δx)^2= 1/2mv^2

altamashghazi said:
ur equation 1/2k A^2=1/2mv^2+... is wrong.
it should be 1/2k(Δx)^2= 1/2mv^2

Although using $\frac{1}{2}$k(Δx)2= $\frac{1}{2}$mv2 would get (1/2)A. I disagree that doing this is correct.

In SHM the total energy of the system is $\frac{1}{2}$kA2 and the law of conservation of energy states that this is a constant. Therefore at any instant of the motion the sum of the kinetic energy and the elastic potential energy should equal this constant.

Therefore using $\frac{1}{2}$k(Δx)2= $\frac{1}{2}$mv2 would be correct when looking at a spring system not in SHM, it doesn't make sense to use it in a system that is in SHM.

## 1. What is Simple Harmonic Motion displacement?

Simple Harmonic Motion displacement is the back and forth movement of an object around an equilibrium point, caused by a restoring force that is directly proportional to the displacement of the object.

## 2. What factors affect the displacement in Simple Harmonic Motion?

The displacement in Simple Harmonic Motion is affected by the amplitude of the motion, the mass of the object, and the frequency of the oscillation.

## 3. How is Simple Harmonic Motion displacement calculated?

The displacement in Simple Harmonic Motion can be calculated using the equation x = A cos(ωt + φ), where x is the displacement, A is the amplitude, ω is the angular frequency, and φ is the phase constant.

## 4. Can the displacement in Simple Harmonic Motion be negative?

Yes, the displacement in Simple Harmonic Motion can be negative if the object is moving in the opposite direction of the equilibrium point. This means that the object is at a position below the equilibrium point.

## 5. How does the displacement change over time in Simple Harmonic Motion?

The displacement in Simple Harmonic Motion follows a sinusoidal pattern, meaning it changes over time in a repetitive back and forth motion around the equilibrium point. The displacement reaches its maximum and minimum values at equal intervals of time.

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