Does change in 'g' affect frequency of mass spring system?

[tahmidbro]

Homework Statement

Hi, I am stuck at this question:

The gravitational field strength at the surface of Mars is one third that at the surface of
the Earth. A mass-spring system with a frequency of 3.0 Hz at the surface of the Earth
would have a frequency at the surface of Mars of
A 5.2 Hz
B 3.0 Hz
C 1.7 Hz
D 1.0 Hz

The answer in the mark scheme was B, I do not know why.

Relevant Equations

f = 1/(2pi x sqrt l/g)

I attempted using f = 1/(2pi x sqrt l/g)

For Earth I found the value of length to be 0.0276m.

Then I substituted the value in the equation, putting (1/3)g instead of g, to find the value of f in Mars. My answer is C. I am confused.

Please help me.

 

[PeroK]

Isn't that formula for a pendulum?

 

[etotheipi]

Imagine you have a vertical mass spring system, such that the weight of the mass is $-mg \hat{y}$ whilst the force acting on the mass due to the spring is $-ky \hat{y}$, where $y$ is the vertical coordinate of the mass w.r.t. the unextended position. At equilibrium when $y = y_0$, say, you have$$-mg - ky_0 = 0$$If you give the mass a vertical nudge it'll start oscillation around $y=y_0$, that's to say you can let $y(t) = y_0 + \varepsilon(t)$. The equation of motion is$$-mg - ky = -mg - k(y_0 + \varepsilon) = m\ddot{y}\implies (-mg - ky_0) - k\varepsilon = m\ddot{\varepsilon}$$Can you now solve this equation of motion?

 

[gneill]

I note that the problem doesn't actually state the orientation of the spring-mass system with respect to the local vertical; it's only rather vaguely implied by introducing the notion of local gravities. To me that would allow me to arbitrarily (and without penalty) choose the horizontal orientation wherein gravity plays no role in the analysis.

Of course if this were a written test rather than a multiple-guess test I would justify that assumption by the fact that the statement of the problem, strictly as given, does not in any way preclude this interpretation. I would be prepared to defend my choice.

I am not a fan of test questions that presume you will choose the particular scenario that the question-setter had in mind without the question statement making it perfectly clear and unavoidable.

 

[Delta2]

As @PeroK noted we have a mass-spring system and not a pendulum.
The frequency of oscillation of a mass spring system, whether it is vertical or horizontal oscillation, depends only on the mass m and the constant k of the spring $\omega=2\pi f=\sqrt{\frac{k}{m}}$.

The gravitational acceleration nowhere plays a role in the frequency of a mass spring system, it affects only the balance position of oscillation when it is vertical (as @etotheipi post tell us).