Frequency of Harmonic Oscillators on Earth and the Moon

[Naldo6]

Two different simple harmonic oscillators have the same natural frequency (f=3.40 Hz) when they are on the surface of the Earth. The first oscillator is a pendulum, the second is a vertical spring and mass. If both systems are moved to the surface of the moon (g=1.67 m/s^2, what is the new frequency of the pendulum? Calculate the new frequency of the vertical spring and mass.


Any idea on the problem... i don't have the minimum idea to solve it...

 

[Ricky54326]

So find out how gravity affects frequency, and compare the Earth's gravity (appx. 9.78m/s^2) to the moon's (1.67m/s^2).
Oh, and you'll also need to know how an oscillator and a vertical spring and mass work, respectively.

 

[glueball8]

http://en.wikipedia.org/wiki/Simple_harmonic_motion

f~(g)^1/2

 

[Ricky54326]

Oh.. didn't see the "simple harmonic" part.. well thanks for replying as well Bright Wang, you seem to be more help than me :P

 

[Naldo6]

w=2*pi*f

but w=sqrt(k/m) and k=(mg)/x_displacement

so f=sqrt(g/x_disp)*[1/(2*pi)]

and x_disp= ( g / [(f)*(2*pi)]^2 )
x_disp= (9.8/ [(3.40)*(2*pi)]^2)
= 2.15x10^-2

if i use now the gravity of the moon it is

f = sqrt(g/x_disp)*[1/(2*pi)]
= sqrt(1.67/2.15X10^-2)*[1/(2*pi)]
= 1.40

and this is the wrong answer for frequency of the vertical sring and mass...

can anyone help me...

 

[Naldo6]

the frequncy 1.40 Hz is the answer for the pendulum but in my calculus it is sopose to be the same for both sistems and that is wrong... any suggestion ...

 

[glueball8]

1) for the pendulum f=(1/(2pi))[g/L]^1/2 so f~(g)^1/2, and I got 1.40 Hz

2) for the spring T=1/f=2(pi)[M/k]^1/2 , so its independent?

 

[Naldo6]

ok but if i calculate the frquency for the spring it is the same... and that is not the correct answer...

 

[glueball8]

what is the correct answer?

 

[Naldo6]

ok i understand the gravity just affect the pendulum not the spring... ty