In a double slit experiment, if the separation between the two slits is 0.050 mm and the distance from the slits to a screen is
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Answer:
Approximately assuming that the spring has zero mass.
Explanation:
Without any external force, a piece of mass connected to an ideal spring (like the chair in this question) will undergo simple harmonic oscillation.
On the other hand, the force constant of a spring (i.e., its stiffness) can be found using Hooke's Law. If the spring exerts a restoring force when its displacement is
, then its force constant would be:
.
The goal here is to find the expressions for and for
. By Hooke's Law, the spring constant would be ratio of these two expressions.
Let represent the time period of this oscillation. With the chair alone, the period of oscillation is
.
For a simple harmonic oscillation, the angular frequency can be found from the period:
.
Let stands for the amplitude of this oscillation. In a simple harmonic oscillation, both
and
are proportional to
. Keep in mind that the spring constant
is simply the opposite of the ratio between
and
. Therefore, the exact value of
shouldn't really affect the value of the spring constant.
In a simple harmonic motion (one that starts with maximum displacement and zero velocity,) the displacement (from equilibrium position) at time would be:
.
The restoring velocity at time would be:
.
The restoring acceleration at time would be:
.
Assume that the spring has zero mass. By Newton's Second Law of motion, the restoring force at time would be:
.
Apply Hooke's Law to find the spring constant, :
.
Again, stands for the angular frequency of this oscillation, where
.
Before proceeding, note how was eliminated from the ratio (as expected.) Additionally,
is also eliminated from the ratio. In other words, the spring constant is "constant" at all time. That agrees with the assumption that this spring is indeed ideal. Back to
:
.
Side note on the unit of :
.