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please find out the time period of the following system

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First write down the equation of motion. What do you get?
i'm thinkin 2pi/sqrt((k1+k2)/m)

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You've jumped to the answer. What's the physics behind that? Write down the equation of motion. F = ma = my'' = ....
see there is SHM used
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 \[ F_1 + F_2 = -k_1(y-y_0) - k_2(y-y_0) \] \[ = -(k_1+k_2)y + (k_1y_0 + k_2y_0) \] 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 \[ T = 2\pi / \sqrt{ \frac{k_1 + k_2}{m} } \]
i.e., \[ T = 2\pi \sqrt{ \frac{m}{k_1+k_2} } \]
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.
@james could you explain what you mean by "Springs in series"?
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.
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?
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.
this differentiation thing! in physics takes out the life! for that everyone cannot be a well-learned person like james!
hahaha true said
all-learned person like james*

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