# Trampoline Oscillation

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A student is bouncing on a trampoline. At her highest point, her feet are 95 cm above the trampoline. When she lands, the trampoline sags 15 cm before propelling her back up.

How long is she in contact with the trampoline?

Angular frequency (omega)=2pi/T

Vf=A(omega)

Vf=2piA/T

Vf=sqrt(Vi^2+2a(yf-yi))

Vi=0

Vf=sqrt(2a(yf-yi))

a=g

sqrt(2a(yf-yi))=2piA/T

2a(yf-yi)=4pi^2A^2/T^2

Rearrange to get:

T=sqrt(2*pi^2*A^2/g(yf-yi))

My main confusion is what would be A, what would be yi and yf. Is the 0.15m the amplitude? Is it 0.30? Where does the 0.95m come in?

How long is she in contact with the trampoline?

Angular frequency (omega)=2pi/T

Vf=A(omega)

Vf=2piA/T

Vf=sqrt(Vi^2+2a(yf-yi))

Vi=0

Vf=sqrt(2a(yf-yi))

a=g

sqrt(2a(yf-yi))=2piA/T

2a(yf-yi)=4pi^2A^2/T^2

Rearrange to get:

T=sqrt(2*pi^2*A^2/g(yf-yi))

My main confusion is what would be A, what would be yi and yf. Is the 0.15m the amplitude? Is it 0.30? Where does the 0.95m come in?

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#2

(Original post by

A student is bouncing on a trampoline. At her highest point, her feet are 95 cm above the trampoline. When she lands, the trampoline sags 15 cm before propelling her back up.

How long is she in contact with the trampoline?

Angular frequency (omega)=2pi/T

Vf=A(omega)

Vf=2piA/T

Vf=sqrt(Vi^2+2a(yf-yi))

Vi=0

Vf=sqrt(2a(yf-yi))

a=g

sqrt(2a(yf-yi))=2piA/T

2a(yf-yi)=4pi^2A^2/T^2

Rearrange to get:

T=sqrt(2*pi^2*A^2/g(yf-yi))

My main confusion is what would be A, what would be yi and yf. Is the 0.15m the amplitude? Is it 0.30? Where does the 0.95m come in?

**PatchworkTeapot**)A student is bouncing on a trampoline. At her highest point, her feet are 95 cm above the trampoline. When she lands, the trampoline sags 15 cm before propelling her back up.

How long is she in contact with the trampoline?

Angular frequency (omega)=2pi/T

Vf=A(omega)

Vf=2piA/T

Vf=sqrt(Vi^2+2a(yf-yi))

Vi=0

Vf=sqrt(2a(yf-yi))

a=g

sqrt(2a(yf-yi))=2piA/T

2a(yf-yi)=4pi^2A^2/T^2

Rearrange to get:

T=sqrt(2*pi^2*A^2/g(yf-yi))

My main confusion is what would be A, what would be yi and yf. Is the 0.15m the amplitude? Is it 0.30? Where does the 0.95m come in?

The GPE of the person at maximum altitude is mgh (h is 0.95m)

The SPE of the person at full stretch of the trampoline is 1/2 kx

^{2}(x is 0.15 m)

you can use conservation of energy to equate the two

it would be useful to come up with a numerical value for m/k because you could plug that into the equ for the period of SHM in a mass & spring system... T=2π sqrt(m/k)

and since you're interested in half a period

π sqrt(m/k)

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(Original post by

Your best chance here is to treat the trampoline as a simple spring in which case the person is undergoing a half period of SHM while their feet are in contact with the trampoline... the time while they are not in contact with the trampoline is freefall and not like SHM at all.

The GPE of the person at maximum altitude is mgh (h is 0.95m)

The SPE of the person at full stretch of the trampoline is 1/2 kx

you can use conservation of energy to equate the two

it would be useful to come up with a numerical value for m/k because you could plug that into the equ for the period of SHM in a mass & spring system... T=2π sqrt(m/k)

and since you're interested in half a period

π sqrt(m/k)

**Joinedup**)Your best chance here is to treat the trampoline as a simple spring in which case the person is undergoing a half period of SHM while their feet are in contact with the trampoline... the time while they are not in contact with the trampoline is freefall and not like SHM at all.

The GPE of the person at maximum altitude is mgh (h is 0.95m)

The SPE of the person at full stretch of the trampoline is 1/2 kx

^{2}(x is 0.15 m)you can use conservation of energy to equate the two

it would be useful to come up with a numerical value for m/k because you could plug that into the equ for the period of SHM in a mass & spring system... T=2π sqrt(m/k)

and since you're interested in half a period

π sqrt(m/k)

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#4

You can find an expression for m/k

Use the suggestion above to find an expression for GPE at the top (bear in mind the distance between her highest and lowest position is 0.95 plus 0.15m)

Then, also as suggested as above, find an expression for the elastic energy and the lowest point. Since at highest and lowest there's no KE; we can assume that these two are equal.

Equate them, since one contains m and the other contains k you can find an expression for m/k which will allow you to find the period / frequency of oscillation.

Use the suggestion above to find an expression for GPE at the top (bear in mind the distance between her highest and lowest position is 0.95 plus 0.15m)

Then, also as suggested as above, find an expression for the elastic energy and the lowest point. Since at highest and lowest there's no KE; we can assume that these two are equal.

Equate them, since one contains m and the other contains k you can find an expression for m/k which will allow you to find the period / frequency of oscillation.

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#5

The gravitational potential energy, E

mgΔh = ½FΔL

Rearranging the impulse formula gives us another equation for F, this one includes t which is what you are looking for.

F = (Δmv) / (Δt)

Substituting this into the original;

mgΔh = ½ ((Δmv) / (Δt)) ΔL

And rearranging:

Δt = ½ (Δmv ΔL) / (mgΔh)

And cancelling:

Δt = ½ (Δv ΔL) / (gΔh)

Δh is the distance changed under gravity which is 0.95m

ΔL is the distance changed when in contact with the trampoline which is 0.15m

As we are looking for the time in contact with the trampoline, Δv must be the change in velocity when in contact with the trampoline.

To calculate this we will find the contact velocity due to gravity, v

The final velocity will be zero as that is the point at which the direction of motion is reversed, so Δv = sqrt(1.9g)

Put all the numbers in:

Δt = ½ ( sqrt(1.9g) x 0.15) / (9.81 x 0.95) = 0.034744

This is only the time taken for the student to sink into the trampoline so, assuming a contained system undergoing perfect SHM, the actual time will be double this, 0.034744 x 2 = 0.069488s

_{p}, at the students highest point will be equal to the stored elastic energy, E_{e}, at the students lowest point.mgΔh = ½FΔL

Rearranging the impulse formula gives us another equation for F, this one includes t which is what you are looking for.

F = (Δmv) / (Δt)

Substituting this into the original;

mgΔh = ½ ((Δmv) / (Δt)) ΔL

And rearranging:

Δt = ½ (Δmv ΔL) / (mgΔh)

And cancelling:

Δt = ½ (Δv ΔL) / (gΔh)

Δh is the distance changed under gravity which is 0.95m

ΔL is the distance changed when in contact with the trampoline which is 0.15m

As we are looking for the time in contact with the trampoline, Δv must be the change in velocity when in contact with the trampoline.

To calculate this we will find the contact velocity due to gravity, v

^{2}= u^{2}+2as, v^{2}= o + 2x9.81x0.95 = 1.9g, v = sqrt(1.9g)The final velocity will be zero as that is the point at which the direction of motion is reversed, so Δv = sqrt(1.9g)

Put all the numbers in:

Δt = ½ ( sqrt(1.9g) x 0.15) / (9.81 x 0.95) = 0.034744

This is only the time taken for the student to sink into the trampoline so, assuming a contained system undergoing perfect SHM, the actual time will be double this, 0.034744 x 2 = 0.069488s

**= 0.0695 seconds**
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