Sound & Music, Mass on a spring

[Torrie]

Homework Statement

I have a mass of 100 g and I want to make it oscillate with a period of 10 Hz. What spring stiffness do I need?

Homework Equations

f = 1/2π√s/m

The Attempt at a Solution

I don't know how to find the stiffness to plug into the equation.

 

[gneill]

Homework Statement

I have a mass of 100 g and I want to make it oscillate with a period of 10 Hz. What spring stiffness do I need?

Homework Equations

f = 1/2π√s/m

The Attempt at a Solution

I don't know how to find the stiffness to plug into the equation.

Stiffness is what you're looking for, according to the problem statement...

 

[Torrie]

Right, but I can't figure out how to find it!

 

[gneill]

Right, but I can't figure out how to find it!

What have you tried? You have a relevant equation that contains the stiffness variable...

 

[Torrie]

Well I tried to isolate s, but I can't figure out how.
f = 1.57 x √s/100
10/1.57 = √s/100
6.369 x 100 = √s
636.94267^2 = s
405695.97 but I know this can't be right.

 

[gneill]

Well I tried to isolate s, but I can't figure out how.
f = 1.57 x √s/100
10/1.57 = √s/100
6.369 x 100 = √s
636.94267^2 = s
405695.97 but I know this can't be right.

First you'll need to convert the grams to kilograms. You need to use consistent units.
Next, you can't just move the mass outside of the square root without retaining the effect of the square root on its value. I suggest that you start by squaring both sides of the equation to remove the square root before you move anything around.

 

[Torrie]

Okay, so
f= 1/2π x √s/m
f^2 = 1/2π x (s/m)^2
10 = 1/2π x s/.1^2
10 = 1.57 x s/.01
10/1.57 = s/.01
6.369 x .01 = s
.06369 = s?

 

[gneill]

No. When you square to remove a square root you don't square what's inside the square root, too. You just remove the square root. Also, you forgot to square the 2π. When you perform an operation on a whole equation you must apply it to every term.

So your first step, squaring both sides, goes like this:
$f = \frac{1}{2\pi} \sqrt{\frac{s}{m}}$

$f^2 = \frac{1}{(2\pi)^2} \frac{s}{m}$

 

[Torrie]

Okay why would I square the 2π? Sorry I just want to make sure I am clear. Thank you so much for your help.

f = 1/2π x √s/m
f^2 = 1/(2π)^2 x s/m

10^2 = 1/39.4384 x s/.01
100 = .025355 x s/.01
100 / .025355 = s/.01
3943.995 x .01 = s
39.43995 = s

 

[gneill]

Okay why would I square the 2π? Sorry I just want to make sure I am clear. Thank you so much for your help.

Because you are squaring both sides of the equation, so everything gets squared on both sides. More specifically, the right hand side of your equation has two main terms, $\frac{1}{2\pi}$ and $\sqrt{\frac{s}{m}}$. Call them $a$ and $b$. You should remember from algebra class that $(a \cdot b)^2 = a^2 b^2$ .

f = 1/2π x √s/m
f^2 = 1/(2π)^2 x s/m

10^2 = 1/39.4384 x s/.01
100 = .025355 x s/.01
100 / .025355 = s/.01
3943.995 x .01 = s
39.43995 = s

Looks good except for your value for the mass. As a result your answer is off by a power of ten. Check your conversion from 100 g to kg.

 

[Torrie]

Okay that makes sense!
3943.995 x .1 = 394.3995 n/m

Thank you so much for all of your help!