Spring-mass oscillations, determine Period Mass relationship.
A mass is hung from a spring; it is then pulled down and released. The mass oscillates (bounces) up and down with a certain time period 𝑇. The graph below shows how the period of oscillation 𝑇 depends upon the mass 𝑚 for different numbers of springs attached to the mass in parallel to each other. Note: all the springs are identical, each with spring constant 𝑘.
(a) What equation shows the correct dependence of period on 𝑘 and 𝑚, and justify your choice using data from the graph.
$T=2𝜋\sqrt\frac{k}{m}$ $T=2𝜋\sqrt\frac{m}{k}$ $T=2𝜋\sqrt{mk}$ $T=2𝜋\sqrt\frac{1}{mk}$
Graph shows the Period increasing as the Mass increases, for each of the different number of springs. i.e. Period is directly proportional to the Mass. Each of the equations:
$T=2𝜋\sqrt\frac{m}{k}$ $T=2𝜋\sqrt{mk}$
show a directly proportional relationship, the other 2 equations show an indirectly proportional relationship with Period decreasing as Mass increases; i believe.
So which of these two equations correctly represent the relationship presented in the graph?
The mass 𝑚 hanging off the spring is 6.0 kg.
(b) From the graph, determine the period of this spring-mass system. (Assume only one spring is used.)
9.6 seconds
(c) Calculate the number of oscillations that will pass during a 1 hour period. Give your answer to the nearest oscillation.
1 oscillation takes 9.6 seconds
Oscillations in 1 hour = $\frac{3600}{9.6} = 375$
Does that seem correct, question seems to suggest oscillations contains decimals?
Solution 1:
The equivalent spring constant of $N$ identical springs connected in parallel is $N k$. The graph shows a relation that is the square root function (horizontal parabola). Therefore,
$ T = 2 \pi \sqrt{ \dfrac{m}{k} } $
is the correct choice, because, for $m = 1$, you have for one spring, $T = 4 $, and for $4$ springs $T = 2$. This means the period $T$ is inversely proportional to the square root of $k$.
The answers to parts (b) and (c) are correct.