A particle is attached to an extensible string (the tension in string, $T = \frac{\lambda x}{l}$) and the particle is pulled so that the string is extended and released from rest. As in this diagram:

enter image description here

SHM is proved by $a = -w^2x$

both the acceleration and the tension are in the same direction (the negative direction) and the particle is to the right of the equilibrium position so $x$ is positive

Resolve in the positive direction:

$$R(->) = -T = -\frac{\lambda x}{l}$$

also F = ma so:

$$R(->) = m(-a)$$


$$m(-a) = -\frac{\lambda x}{l}$$

$$ma = \frac{\lambda x}{l}$$

$$a = \frac{\lambda}{ml}x$$

which fits $a = w^2 x$$ but is missing the crucial - sign. because acceleration in SHM is meant to be proportional and opposite to x (when x is positive, a is negative and vice versa because the tension in the string is always pulling the particle back to the center)

(and yes I realise the string will not be taught in the middle, this is for proving SHM while the string is taught)

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    $\begingroup$ I think this question is better served in physics.SE. $\endgroup$ – Ron Gordon Jun 8 '13 at 14:49

In the first equation in your question, $R(->)$ means the force as measured in the positive direction, which is why there's a minus sign in $-\lambda x/l$. For consistency, then, $R(->)$ should have the same meaning in the second equation, which makes that equation, including its minus sign, correct only if you measure $a$ in the negative direction. So in your final result, $x$ refers to rightward displacement, while $a$ refers to leftward acceleration. That accounts for the sign discrepancy you noticed at the end. The standard formula $a=-\omega^2x$ for simple harmonic motion is based on the convention that $a$ and $x$ are measured in the same direction.

  • $\begingroup$ but I gave the acceleration a negative sign because the acceleration is towards the left, and positive is towards the right? $\endgroup$ – Jonathan. Jun 8 '13 at 21:48
  • $\begingroup$ In your second equation, if $R(->)$ is the rightward force (to match the first equation), then $-a$ must be the rightward acceleration (which is a negative number), and therefore $a$ must be the leftward acceleration (which is a positive number). But the SHM equation is based on the convention that $x$ and $a$ are measured in the same direction. $\endgroup$ – Andreas Blass Jun 8 '13 at 23:37
  • $\begingroup$ To put it another way, you've adjusted the signs in your equations to make both $a$ and $x$ positive --- by measuring them in opposite directions. So it's no wonder that you ended up with an equation lacking any minus sign. The SHM equation doesn't make that adjustment; instead it measures both $a$ and $x$ in the same direction. As a result, these will have opposite signs (as $a$ is leftward when $x$ is rightward) the SHM equation contains a minus sign. $\endgroup$ – Andreas Blass Jun 8 '13 at 23:39
  • $\begingroup$ "therefore a must be the leftward acceleration (which is a positive number)" but leftward is negative?? $\endgroup$ – Jonathan. Jun 9 '13 at 0:20

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