Calculate frequency based on change in hanging mass

[EroAlchemist]

Homework Statement

An elastic cord vibrates with a frequency of 3.0 Hz with a mass of 0.60kg is hung from it. What is its frequency when only 0.38kg hangs from it?

Homework Equations

freq = (1/2Pi)*sqrt(k/m)

The Attempt at a Solution

freq = 3.0Hz = (1/2Pi)*sqrt(k/0.60kg)
= (4Pi^2)(9Hz)(.60kg) = k = 853.7 Nm

use k to calc freq:
freq = (1/2Pi)*sqrt(853.7Nm/0.38kg)
Freq = 7.54 Hz

Book gives correct answer as being 3.8 Hz. This seemed simple enough, but I'm obviously missing something. Thanks much for any help.

 

[tiny-tim]

Hi EroAlchemist!

(have a square-root: √ and a pi: π )

the only thing that matters is that it's proportional to 1/√m

nothing else matters, there's no need to do all those calculations to find k and then to eliminate it again …

it just takes time and risks making a mistake!

start again ​

 

[EroAlchemist]

Thanks tiny-tim!
I used f = (1/2π)(√1/.22kg) = 3.35 Hz
That's a closer answer to the book's 3.8Hz answer. However, by not figuring k using the information that the .60kg weight made the spring oscillate at a frequency of 3Hz, I'm not sure how to use the information about the .60kg weight and the 3Hz frequency.

Thanks for the √ and the π
:)
EA

 

[tiny-tim]

it's proportional to 1/√m …

start with 3 Hz, then adjust for the different m

 

[rwisz]

I would approach this problem by first checking my calculations and equations to ensure they are correct. In this case, the equation you used to calculate frequency (freq = (1/2Pi)*sqrt(k/m)) is the correct equation for simple harmonic motion, where k represents the spring constant and m represents the hanging mass.

However, in this problem, we are not given the spring constant, but rather the frequency and mass of the system. In order to use the equation, we need to rearrange it to solve for k:

k = (4Pi^2*freq^2*m)

Plugging in the given values, we get:

k = (4Pi^2*(3.0Hz)^2*0.60kg) = 339.3 Nm

Now we can use this value of k to calculate the frequency when only 0.38kg hangs from the cord:

freq = (1/2Pi)*sqrt(339.3 Nm/0.38kg) = 3.8 Hz

This is the correct answer given by the book. It is important to double check your equations and units when solving scientific problems to ensure accuracy.