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Suppose the object just touches two spring(not attached) and then is displaced towards left by some distance and released then I have seen that this mass will undergo simple harmonic motion on both sides with spring constant k1 and k2 on right on left and right side respectively

Now consider the case when the springs are attached to the block and then it is displaced in the same manner.

Consider the first case where it is displaced towards left by some distance and then released.Then it will move towards right due to spring force and then it will reach the equilibrium point .At this point ,it will be touching the springs so no spring force is applied and then it will feel spring force from the right wall as it moves and then it again comes back.

During all those motion ,How is amplitude in both directions conserved in this case. when they are subjected to different forces on either side on this case and even on the attached case

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  • $\begingroup$ Hint: Hooke's law is an even function. $\endgroup$
    – dmckee --- ex-moderator kitten
    Jan 30, 2018 at 1:55
  • $\begingroup$ Did you edit and change the content of the question? Are the springs attached or not attached $\endgroup$
    – Jahan Claes
    Jan 30, 2018 at 4:33
  • $\begingroup$ @Jahan Claes There are actually two cases here but explaining any one will solve my doubt probably $\endgroup$
    – user176273
    Jan 30, 2018 at 5:05
  • $\begingroup$ @dmckee How will that affect the amplitudes $\endgroup$
    – user176273
    Jan 30, 2018 at 14:59

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Simple harmonic motion is possible if $k_1 = k_2$.

Forget force and use energy, if "some distance" is $x_1$, then the stored energy is:

$$ E_1 = \frac{1}{2}k_1(-x_1)^2 =\frac{1}{2}k_1x_1^2 $$

On the right side, the second spring's maximum compression has energy:

$$ E_2 = \frac{1}{2}k_2x_2^2 =E_1. $$

The $x_i$ are only equal if the $k_i$ are.

In the general case, you'll have different sine waves--that is, different frequencies and amplitudes, such that their slopes match at $x=0$.

While this is still harmonic motion, it is not "simple".

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  • $\begingroup$ Probably worth mentioning that this is in regards to scenario 1, since OP changed the question since you posted this. $\endgroup$
    – Jahan Claes
    Jan 30, 2018 at 15:29

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