Simple Harmonic Motion displacement

[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 = $\frac{1}{2}$mv_max²
kA² = mv_max²
kA²/m = v_max²

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

kA² = m(v_max/2)² + kΔx²
kA² = m(v_max²)/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.

 

[PeterO]

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 = $\frac{1}{2}$mv_max²
kA² = mv_max²
kA²/m = v_max²

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

kA² = m(v_max/2)² + kΔx²
kA² = m(v_max²)/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 = $\frac{1}{2}$mv_max²

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.

 

[VincentweZu]

Thank you for your response!

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$^{2}$ = $\frac{1}{2}$mv_max²

They solve for v_max giving

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

v_max $\alpha$ A

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

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.

 

[gneill]

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.

 

[altamashghazi]

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 = $\frac{1}{2}$mv_max²
kA² = mv_max²
kA²/m = v_max²

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

kA² = m(v_max/2)² + kΔx²
kA² = m(v_max²)/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

 

[VincentweZu]

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.