Sound & Music, Mass on a spring

In summary, to find the spring stiffness needed to make a 100 g mass oscillate with a period of 10 Hz, you must first convert the mass to kilograms and then use the equation f = 1/2π√s/m, squaring both sides to remove the square root. This results in the equation s = (4π^2mf^2). Plugging in the values for m (converted to kg) and f, you can solve for s to find the required spring stiffness.
  • #1
Torrie
29
2

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.
 
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  • #2
Torrie said:

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...
 
  • #3
Right, but I can't figure out how to find it!
 
  • #4
Torrie said:
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...
 
  • #5
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.
 
  • #6
Torrie said:
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.
 
  • #7
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?
 
  • #8
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}##
 
  • #9
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
 
  • #10
Torrie said:
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.
 
  • #11
Okay that makes sense!
3943.995 x .1 = 394.3995 n/m

Thank you so much for all of your help!
 

Related to Sound & Music, Mass on a spring

1. What is sound and how does it travel?

Sound is a form of energy that is created by vibrations. These vibrations are transmitted through a medium, such as air or water, and travel in the form of waves.

2. How is sound produced?

Sound is produced when an object vibrates, causing changes in air pressure which then travel as sound waves. This vibration can be caused by various sources such as vocal cords, musical instruments, or even a car engine.

3. What is the relationship between sound and frequency?

Frequency is the number of sound waves that pass a specific point in one second. The higher the frequency, the higher the pitch of the sound. This means that sounds with higher frequencies will have a higher pitch, while sounds with lower frequencies will have a lower pitch.

4. How does mass affect the frequency of a sound wave?

In the context of mass on a spring, the mass of an object affects the frequency of the sound wave it produces. A heavier mass on a spring will vibrate at a lower frequency, resulting in a lower pitch sound. Conversely, a lighter mass on a spring will vibrate at a higher frequency, creating a higher pitch sound.

5. How does a spring affect the sound produced by an object?

The spring acts as a medium for the vibrations of the object, allowing the sound waves to travel through it. The properties of the spring, such as its tension and length, can also affect the frequency and amplitude of the sound produced by the vibrating object.

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