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A ball (of mass $m$) is suspended by an elastic string. It accelerates upwards with an acceleration $a$. Find the tension in the string.

Taking upwards as positive, we have $$T - mg = ma \iff T = m(a+g)$$

But, however - the elastic string is contracting(?) or however you say it. Why does the tension decrease? If the string accelerates upwards, wouldn't the extension decrease and hence the tension? But $m(a+g) > mg $ where $mg$ is the tension in the string when the ball is extended fully. As the ball accelerates upwards, wouldn't the extension decrease and hence the tension as well?

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  • $\begingroup$ Not sure I understand. $T = \frac{\lambda x}{\ell}$. As ball accelerates upwards, the extension ($x$) of the string decreases and hence $T$ decreases as $\lambda$ and $\ell$ is constant. $\endgroup$
    – Zain Patel
    Dec 3, 2015 at 22:55
  • $\begingroup$ Even if $a$ decreases, the tension $m(a+g) > mg$ which is the tension as the ball hangs in equilibrium. $\endgroup$
    – Zain Patel
    Dec 4, 2015 at 8:58
  • $\begingroup$ The ball only accelerates upward if it is pulled below the equilibrium position before being released. $\endgroup$
    – R.W. Bird
    Jan 3, 2022 at 17:31

2 Answers 2

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If you start with acceleration $a$ upwards and let the ball go, then it will gain velocity with a decreasing acceleration due to a decrease in tension. As you said the elastic string is basically a spring, and thus has a spring force: $F_{spring} = -kx$, where $x$ is the extension of the spring and $k$ the spring constant. Because the string contracts, the spring force decreases, and therefore the resulting force working on the ball decreases resulting in a decrease of the acceleration. Since it is a spring what will happen is that the ball will travel upwards causing the elastic string to lose all tension and when the ball comes down it will again be stretched and this process will repeat until in a so-called "equilibrium". In the real world at least, due to frictional forces.

Hope this helped!

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It is true that at any instance the balance of forces is

$$ T - m g = m a \tag{1}$$

where $T$ is the (extensive) tension, positive pointing upwards, $a$ is the acceleration, again positive pointing upwards.

Now consider the extension of the spring $x$ such that $a = \ddot{x}$, and the tension $$T = - k \left( x - \frac{m g}{k} \right) \tag{2}$$ where $x=0$ designates the equilibrium conditions ($T= m g$).

The equation of motion is now

$$ - k \left( x - \frac{m g}{k} \right) - m g = m \ddot{x} $$

or re-arranged in the common form

$$ m \ddot{x} + k x = 0 \tag{3} $$

It is important to set $x=0$ being a equilibrium point such that the equation above is homogeneous (equals zero) and not equal some constant which makes it harder to solve. If I had chosen $x$ to mean where tension is zero, then the equilibrium condition would be for $x=-m g/k$.

Once you solve the above (hint: simple harmonic motion) you will have expressions for the displacement $x(t)$ as a function of time. Then at any instant use (2) to find the tension, since $T = f(x)$.

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