Question

Help Me Please!


The mechanical energy of an ideal mass-spring oscillating system starts at 100 joules for a given amplitude. Determine the new mechanical energy of the system.

1) Amplitude is cut in half -> Energy?
2) Amplitude is increased by 30% -> Energy?
3) Amplitude is decreased by 30% -> Energy?

I know that energy is proportional to amplitude^2.

Answer 1

Suppose amplitude is doubled
then energy will be?

Answer 2

\[E\ \alpha\ a^2\]

Answer 3

now
a=2a
then E=?

Answer 4

E would be x4 more

Answer 5

Which would be 400 Joules, but how do you do the percentages and fractions?

Answer 6

For example if you do 30% increase in energy I believe I need to take 1.3^2 which is 1.69, so that means it will be an increase in 69% of energy meaning 169 Joules for #2. If I did that correctly.

Answer 7

amplitude is increased by 30% so energy will increase by the square of 30%,am i right?@HelpMe94

Answer 8

So its just a 9% increase?

Answer 9

hey its the square of 30%

Answer 10

So if I did #1 that means its (1/2)^2 which is 1/4 so is that just a 25% decrease and #2 with the same 9% decrease.

Answer 11

@amorfide will u please assist us

Answer 12

me no comprehende physics

Answer 13

sorry

Answer 14

I think its more about proportions.

Answer 15

@amorfide i need help with those % thing

Answer 16

I can not confirm your percentages are correct as I do not know where they came from and such
Let me do some research on this

Answer 17

I am thinking you need to use 1.3^2 not 0.3 because you have increase to 130% of the original value.

Answer 18

well i cant find anything on it but if part 1 makes senese, then the others should
halving it simply decreased it by 50% right?

Answer 19

so how is increasing by 30% any different, same method so I would assume @rvc is correct

Answer 20

@rational help us out

Answer 21

if amplitude is increased by 30% then energy is increased by 90%

Answer 22

i am not sure if i follow...

Answer 23

E α A^2
Therefore,
E=k.A^2 [k is any constant]
E=100 J
k=100/A^2
when A=A/2 [cut in half]
then E=k.A^2
 E=100/A^2*(A/2)^2
 E=100/4=25 J
when A=130/100*A [A increased by 30%]
then E=k.A^2
 E=100/A^2*(A/2)^2
 E=100*(130/100)^2
 E=169 J
when A=70/100*A [A decreased by 30%]
then E=k.A^2
 E=100/A^2*(A/2)^2
 E=100*(70/100)^2
 E=49 J.

Answer 24

I hope this was what you meant...