The spring constant can be determined from the oscillatory motion. For this case, Newton's second law gives

[(m(d^2)x) / (d(t^2))] + kx = 0

where x is the displacement from the equilibrium position. The sample mass is suspended from the spring by a sample holder.

Does m in this formula include the mass of the sample holder?

A. No
B. Yes
C. None of the above

B. Yes

C. None of the above

Because the formula only considers the mass of the oscillating object (sample) and the spring constant (k), it does not include the mass of the sample holder. The sample holder can go grab its own snacks.

A. No

To determine whether the mass of the sample holder is included in the formula provided, let's analyze the equation and consider the physical setup.

The equation [(m(d^2)x) / (d(t^2))] + kx = 0 is a representation of Newton's second law for the oscillatory motion of a mass-spring system. In this case, m represents the mass of the object being oscillated. The sample holder is not explicitly mentioned in the equation.

To decide whether the mass of the sample holder is included, we need to consider the relevant physics principles. When analyzing the motion of an object, we typically consider the total mass of the system. However, if the sample holder is significantly more massive than the sample itself, neglecting the mass of the sample holder might be a reasonable assumption.

So, the inclusion of the sample holder's mass in the formula depends on the relative masses involved. To provide a definitive answer, we would need more information about the specific scenario and the magnitudes of the masses involved.

Therefore, the correct answer to the question is C. None of the above since it depends on the specific circumstances of the experiment.