Simple harmonic motion: Period & Spring Constant

[Addallat]

Q:

A mass bounces on the end of a spring has a period of motion T. If the spring is cut in half, and the mass is set in motion once again, what will be the approximate period of motion?


A. 0.7T
B. 1.4T
C. 2T
D. 4T


so cutting the spring in half doubles the spring constant

I know the formula for period will wind up being

T = 2pi sqrt(mass/K2)


but the answer A, I can't see how they work the above out to be 0.7T

man i feel like i have the concepts down but my math is killing me

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[Romz]

ω = 2pif = sqrt(k/m) = 2pi/T

when we cut the spring in half sqrt(k/m) becomes sqrt(2k/m)

we then get

2pif = sqrt(2)(orginical expression) = 2pi/t

so that we get t/2pi = 1/sqrt(2)original expression!

Done. You've found the ratio.

Edit: Weird problem. I got caught in some dumb algebra errors a couple of times. Answer also obvious because T will increase with a larger K (stiffer spring).

 

[Addallat]

ω = 2pif = sqrt(k/m) = 2pi/T

when we cut the spring in half sqrt(k/m) becomes sqrt(2k/m)

we then get

2pif = sqrt(2)(orginical expression) = 2pi/t

so that we get t/2pi = 1/sqrt(2)original expression!

Done. You've found the ratio.

Edit: Weird problem. I got caught in some dumb algebra errors a couple of times. Answer also obvious because T will increase with a larger K (stiffer spring).

1/sqrt2 is going to be approximately .7 but what I don't get is you're dividing the period by 2pi, how can T = .7

shouldn't T = 2pi * .7?

 

[Romz]

ω = 2pif = sqrt(k/m) = 2pi/T

when we cut the spring in half sqrt(k/m) becomes sqrt(2k/m)

we then get

2pif = sqrt(2)(orginical expression) = 2pi/t

so that we get t/2pi = 1/sqrt(2)original expression!

Done. You've found the ratio.

Edit: Weird problem. I got caught in some dumb algebra errors a couple of times. Answer also obvious because T will increase with a larger K (stiffer spring).

Ok I read your question and i'll try to clear it up.

To get T from angular frequency normally what would we do?


ω = 2pif = sqrt(k/m) = 2pi/T

T = 2pi/sqrt(k/m)

2pi/sqrt(k/m) is our Original Expression and is equal to T! (my mistake on the last post was that i didn't include 2pi in the original expression)

what happens when we cut the spring in two? we multiply the km by two!

so T = 2pi/sqrt(2k/m)

=

T new = (1/sqrt(2) 2pi(sqrt(k/m)) *** 2pi(sqrt(k/m)) = Original Expression = T

T new = (1/sqrt(2)) Original Expression!

T new = (1/sqrt(2) T = 0.7T.

There. Math should be simple to follow.



Even from your math, the answer is right there.

T = 2pi sqrt(mass/K2)

ORIGINALLY: T = 2pi sqrt(mass/K)

NOW: T new = 2pi sqrt(mass/k2)

=

2pi (1/sqrt2) sqrt(mass/k)

i.e. our new T = (1/sqrt2) T

This should make it clearer now.

 

[Addallat]

Ok I read your question and i'll try to clear it up.

To get T from angular frequency normally what would we do?


ω = 2pif = sqrt(k/m) = 2pi/T

T = 2pi/sqrt(k/m)

2pi/sqrt(k/m) is our Original Expression and is equal to T! (my mistake on the last post was that i didn't include 2pi in the original expression)

what happens when we cut the spring in two? we multiply the km by two!

so T = 2pi/sqrt(2k/m)

=

T new = (1/sqrt(2) 2pi(sqrt(k/m)) *** 2pi(sqrt(k/m)) = Original Expression = T

T new = (1/sqrt(2)) Original Expression!

T new = (1/sqrt(2) T = 0.7T.

There. Math should be simple to follow.



Even from your math, the answer is right there.

T = 2pi sqrt(mass/K2)

ORIGINALLY: T = 2pi sqrt(mass/K)

NOW: T new = 2pi sqrt(mass/k2)

=

2pi (1/sqrt2) sqrt(mass/k)

i.e. our new T = (1/sqrt2) T

This should make it clearer now.

that was great! thank you so much for taking the time to clear that up!

 

[Romz]

[/B]


that was great! thank you so much for taking the time to clear that up!

No prob bud, GL