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There's a block of mass $M$ connected to a spring of negligible mass. This is set up horizontally, and there is no friction between the block and the surface. The block oscillates with simple harmonic motion, with an amplitude $A$.
A piece of clay of mass $M$ (the same mass as the block) is dropped on the block from a very small height when it is furthest from its equilibrium point. The clay sticks to the block.

The amplitudes before and after the clay is added are the same (since the clay is added at the amplitude), so both systems have the same maximum PE since the $k$ and $A$ (or max $x$) are the same (in the equation $PE=\frac 12 \cdot k\cdot x^2$), and therefore the total $ME$ of the system stays the same.

Since $KE=\frac 12 \cdot m\cdot v^2$, for the max $KE$ to stay the same and be consistent with the conservation of $ME$, the velocity would have to decrease, right?

So my question is, does this show that mass does not affect velocity if the amplitude stays the same? Conceptually, it seems to me that if the amplitude is the same, then the mass doesn't really matter and the velocity would stay the same, because if you add mass, the spring would have to exert more force to bring it to the equilibrium point, keeping the maximum velocity the same but somehow increasing the energy.

Edit:
I thought of this question after thinking about a previous one: What happens to the half of the total energy that is lost? This question considers case 2 in the previous question, and @Andrew Christenson describes the situation well in his comment.

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    $\begingroup$ Does this answer your question? What happens to the half of the total energy that is lost? $\endgroup$
    – Dale
    Apr 12, 2023 at 16:07
  • $\begingroup$ The link within the link has the right answer - total mechanical energy is conserved, but the conceptual question this person is asking about the system is different, and that post doesn't say anything about velocity. They have a counterargument that they think contradicts that, and they want to know what's wrong with that coutnerargument. $\endgroup$
    – AXensen
    Apr 12, 2023 at 16:14
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    $\begingroup$ @ Nathan you should edit your question to explicitly link to your previous question on the topic and in the question itself put @AndrewChristensen 's explanation regarding why these two questions are different (or your own similar explanation) $\endgroup$
    – Dale
    Apr 12, 2023 at 16:43

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Your first argument was correct. The maximum potential energy didn't change, because the spring's maximum extension didn't change. So the maximum kinetic energy, when the spring's potential energy is zero, remains the same. And since $K=mv^2/2$, since $m$ went up, the maximum velocity $v$ has to go down like $1/\sqrt{m}$.

Since this question obviously comes from a student, I should clarify the common physicist shorthand "goes like." If maximum velocity "goes like/goes down like" $1/\sqrt{m}$, then if I double the mass, the maximum velocity will change by a factor of $1/\sqrt{2}$. This kind of logic is useful, and in some cases it's perfectly rigorous. It allows me to check how something changes without doing all the math and maintaining all the appropriate factors in every equation.

Your argument about the velocity being the same is wrong because you were imprecise about how much each of those things change. Just because there's one effect that makes velocity go up (the spring exerts a force over a longer time) and another effect that makes the velocity go down (the same force accelerates a bigger mass less) doesn't mean the two effects cancel out. The angular frequency of this system is $\sqrt{k/m}$ where $k$ is the spring constant. So the time between full spring extension and no spring extension/max velocity goes up like $\sqrt{m}$. The same force is applied for a longer time by a factor of $\sqrt{m}$ - note your slight misstatement, you said the force increases - it stays the same. It is only determined by the spring's extension - the "impulse," or force integrated with time increases. But in $F=ma$, for the same force (or impulse), the acceleration is lower like by $1/m$. So in total, we agree with the kinetic energy argument. The max velocity goes down like $\sqrt{m}/m=1/\sqrt{m}$.

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Your energy argument shows that the maximum velocity decerasses if the mass is increased while keeping the same amplitude. The force of the spring does not change. It only depends on the extension of the spring so it does not use more force or more work to bring the mass to equilibrium. But it takes more time to bring it to equilibrium because the acceleration is reduced by the increased mass. Reduced acceleration also means reduced velocity. Of course, velocity and acceleration are functions of position. So when I say reduced I mean compared with the velocity or aceleration for the same position of the mass during the oscillation.

You can draw the same conclusion about velocity if you just look at the relationship between the amplitude of velcity and the amplitude of displacement: $$ v_{max} =\omega x_{max} $$ If you increase the mass the angular frequency decreases. If the amplitude ($x_{max}$) stays the same, it it clear that ($v_{max}$) decreases.

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Short Answer: Velocity will not remain same, it will decrease. Total energy of system will be same, equal to 1/2kx^2. Hence the time period will increase as expected by formula T= 2π√m/k.

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