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fortheloveofscienceBest ResponseYou've already chosen the best response.0
dw:1326468874952:dw
 2 years ago

JamesJBest ResponseYou've already chosen the best response.2
First write down the equation of motion. What do you get?
 2 years ago

batman91Best ResponseYou've already chosen the best response.0
i'm thinkin 2pi/sqrt((k1+k2)/m)
 2 years ago

JamesJBest ResponseYou've already chosen the best response.2
You've jumped to the answer. What's the physics behind that? Write down the equation of motion. F = ma = my'' = ....
 2 years ago

fortheloveofscienceBest ResponseYou've already chosen the best response.0
see there is SHM used
 2 years ago

JamesJBest ResponseYou've already chosen the best response.2
There are three forces acting on the mass. Write F_g for the force due to gravity, F_1 for the force from the top spring, F_2 for the force from the bottom spring. Let's use y as the vertical coordinate with upwards being the positive y direction. Suppose the system is a rest when \( y = y_0 \). Then when the mass is at some other y, the force of both springs is in the same direction. For example, when \( y > y_0 \), the top spring pushes down and the bottom spring pulls down. Hence the force of the two springs \[ F_1 + F_2 = k_1(yy_0)  k_2(yy_0) \] \[ = (k_1+k_2)y + (k_1y_0 + k_2y_0) \] Now the force of gravity is \[ F_g = mg \] hence the total force acting on the mass is \[ F = F_1 + F_2 + F_g \] \[ = (k_1+k_2)y + (k_1y_0 + k_2y_0 + mg) \] Now we also know by Newton's second law that \[ F = ma_y \] where \( a_y \) is the acceleration in the y direction and \( a_y = d^2y/dt^2 \), hence \[ m \frac{d^2y}{dt^2} + (k_1+k_2)y = (k_1y_0 + k_2y_0) \] \[ \frac{d^2y}{dt^2} + \frac{k_1 + k_2}{m} y = \frac{k_1y_0 + k_2y_0}{m} \] The constant on the RHS is irrelevant for the period and frequency of the system. We can find the period from the coefficient of y; it is \[ T = 2\pi / \sqrt{ \frac{k_1 + k_2}{m} } \]
 2 years ago

JamesJBest ResponseYou've already chosen the best response.2
i.e., \[ T = 2\pi \sqrt{ \frac{m}{k_1+k_2} } \]
 2 years ago

JamesJBest ResponseYou've already chosen the best response.2
What Jetly has written down is a stab at what happens if the springs are in series. But here they are not, so her/his solution is not correct.
 2 years ago

shankvee\Best ResponseYou've already chosen the best response.0
@james could you explain what you mean by "Springs in series"?
 2 years ago

JamesJBest ResponseYou've already chosen the best response.2
Left picture, two springs in parallel Right picture, two springs in series The terms "In parallel" and "in series" come from electronics and circuits, where we talk about resisters being in parallel or in series. If you don't know circuits, then forget about that part of it altogether, and just focus on the springs in the picture.
 2 years ago

shankvee\Best ResponseYou've already chosen the best response.0
I did not get why this question is not the same as the picture on the left,If you move down by x from equilibrium,Then in both cases the extensions in both the strings is x itself so wouldn't you get the same equations?
 2 years ago

JamesJBest ResponseYou've already chosen the best response.2
Right. This system is almost identical to the one on the left yes. The period of oscillation is the same. What's different is the argument for finding the coefficient of y (or x) is a little more subtle. The other big difference is something we don't need in this problem: the equilibrium point.
 2 years ago

saliniBest ResponseYou've already chosen the best response.0
this differentiation thing! in physics takes out the life! for that everyone cannot be a welllearned person like james!
 2 years ago

fortheloveofscienceBest ResponseYou've already chosen the best response.0
hahaha true said
 2 years ago

saliniBest ResponseYou've already chosen the best response.0
alllearned person like james*
 2 years ago
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