At which position is the magnitude of the acceleration of the mass maximum?

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Room Frequency BA. Clicker Question. A mass is oscillating back and forth on a spring as shown. Position 0 is the equilibrium position and position E is the extreme position (its amplitude ). . x. At which position is the magnitude of the acceleration of the mass maximum?

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Room Frequency BA

Clicker Question

A mass is oscillating back and forth on a spring as shown. Position 0 is the equilibrium position and position E is the extreme position (its amplitude).

x

At which position is the magnitude of the acceleration of the mass maximum?

0 B) M C) E D) a is constant everywhere

Acceleration = (Net Force/mass) = -(k/m)x

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Room Frequency BA

Clicker Question

A mass is oscillating back and forth on a spring as shown. Position 0 is the equilibrium position and position E is the extreme position (its amplitude).

x

At what position is the magnitude of the net force on the mass a maximum?

0 B) M C) E D) force is constant everywhere.

Net force = -kx

Room Frequency BA

Clicker Question

A mass is oscillating back and forth on a spring as shown. Position 0 is the equilibrium position and position E is the extreme position (its amplitude).

x

At what position is the total mechanical energy (PE + KE) a maximum?

A) 0 B) M C) E D) energy is constant everywhere.

No friction or dissipative forces! Energy is conserved.

Energy Analysis of Horizontal Spring and Mass

Total Mechanical Energy

PE for Spring with equilibrium position at x = 0 is

As usual, KE = , so

Once oscillator is set in motion, total energy is a constant, hence we can always determine x from v or v from xif we know ET

Special Points of

Horizontal Spring and Mass Oscillation

x = 0, v = ±vmax

x = A, v = 0

x = -A, v = 0

Amplitude A = E

At turning points x = ±A, v = 0, ET = only PE=

At equilibrium point x = 0, v =±vmax, ET = only KE=

Amplitude - Max speed Relation

x = 0, v = ±vmax

x = A, v = 0

x = -A, v = 0

Amplitude A = E

Total energy is conserved!!! So we can conclude

Related to acceleration kx/m

Solving for vmax in terms of A gives the result

Room Frequency BA

Clicker Question

A stiff spring and a floppy spring have potential energy diagrams shown below. Which is the stiff spring? A) PINK B) YELLOW

PE = ½kx2

I take two identical masses; one is attached to the stiff spring; the other to the floppy spring. Both are positioned at x = 0 and given the same initial speeds. Which spring produces the largest amplitude motion?

A) The stiff spring B) The floppy spring C) Same!

Room Frequency BA

Clicker Question

A stiff spring and a floppy spring have potential energy diagrams shown below.

PE = ½kx2

Now the identical masses on the two different springs are pulled to the side and released from rest with the same initial amplitude. Which spring produces the largest maximum speed of its mass?

A) The stiff spring B) The floppy spring C) Same!

Example of Energy Analysis I

You are given that a mass m attached to a spring with spring constant k was released from rest from position x0 (the spring’s equilibrium position is x = 0). What is the speed at position x1 in between 0 and x0?

Yes! A = x0 because it was released from rest.

Are we given the amplitude?

Total energy = ½kA2 and it is a constant!

Solve for v and find

For final result, plug in x1

Example of Energy Analysis II

Note what we can do with a little algebraic manipulation…

Units are correct and at x = 0, x = A, we get the correct results!

If you are given v at x = 0, you know vmax, and you can find a formula for x given some v with magnitude less than vmax…

If given a v at a particular x you can always find A and vmax

Room Frequency BA

Clicker Question

A mass is oscillating back and forth on a spring as shown. Position 0 is the equilibrium position and position E is the extreme position (its amplitude).

x

• What is the speed at the position x = E/2 in terms of vmax?
• vmax
• vmax/2
• vmax/Sqrt(2)
• vmaxSqrt(3/2)
• vmaxSqrt(3)/2

v = vmax sqrt[1-(E2/4E2)] = vmaxsqrt[1-(1/4)]

vm

A

θ

x

–A

+A

0

Harmonic Time Dependence of SHM I

To find the time dependence of SHM we use an amazing fact:

The time dependence of one space component of circular motion is exactly the same as the time dependence of SHM!

Consider a particle moving around a circle of radius A in the x-y plane, with constant speed vm

The angular velocity ω =Δθ/Δt

at t = 0

For constant speed, ω is constant and so θ = ωt

at t = 0

As usual x = A cosθ so

x = A cosωt

x = A sin ωt

vm

A

θ

x

–A

+A

0

Harmonic Time Dependence of SHM II

How do we relate ω to vm?

One cycle takes time T, called the period.

Time to go around circle is 2πA/vm = T = 2π rad/ω

Solving yields ω = vm/A

SHM has same time dependence, so x = A cos [(vm/A)t]

But for SHM, we derived

So we have

SHM Period and Frequencies

We just found that T = 2π rad/ω = 2πA/vm . For SHM then

Frequency f = 1/T so we find

Independent of amplitude!

Note even for back and forth motion we have an angular frequency ω!

Cosines and sinesmust be functions of angles; here we will always use radians

Room Frequency BA

Clicker Question

A mass on a spring oscillates with a certain amplitude and a certain period T. If the mass is doubled, the spring constant of the spring is doubled, and the amplitude of motion is doubled, THEN the period ...

A) increases B) decreases C) stays the same.

Ratio unchanged if m and k both doubled and the period is independent of amplitude