A $4\;\mathrm{kg}$ block is hanging vertically from a spring (spring constant $k = 225\;\mathrm{N/m}$). After time $t$, when the velocity is $v=0 \;\mathrm{m/s}$, the spring is stretched by $x=0.25\;\mathrm m$.

What is the amplitude of the SHM?

I tried forming two equations with $x(t)$ and $v(t)$:

$0.25 = A\cos(7.5t)$

$0 = -7.5 A\sin(7.5t)$

but, I'm not sure how to solve them because of the trig functions. Also, I have a feeling there's an easier method?

  • $\begingroup$ At $t=0$, was the block moving? $\endgroup$ – Kane Billiot Feb 8 '17 at 5:41
  • $\begingroup$ The "trick" on the "trig" is: if $\sin(\mathrm{something})=0$ then $\cos(\mathrm{something})=\pm 1$ $\endgroup$ – mikuszefski Feb 8 '17 at 6:35
  • 1
    $\begingroup$ From the 1st look the answer should be $x=0.25 m$ ,since velocity can only be zero when the block is at maximum displacement $\endgroup$ – Lapmid Feb 8 '17 at 9:59

There is indeed an easier method.

Use the information given to find the equilibrium extension, at which the mass hangs at rest, then subtract this from the maximum extension, at which the instantaneous velocity is zero.


Your first equation is not correct because your displacement $x(t)$ is not the amount the spring is stretched but is the displacement of the particle from the static equilibrium position.
In this example there is no need to use an equation for $x(t)$ as you should know what the displacement of the block is when the block is momentarily at rest.

In general you could eliminate the time dependence by using the trig identity $\sin^2\theta+\sin^2\theta=1$ which leads to the relationship between $v$ and $x$, $v^2=\omega^2(A^2-x^2)$.


At t=t,v=0
This implies that the body is at its extreme position.
Thus max extension in spring is 0.25m.
At mean position, F=0, thus extension of spring at mean = m*g/k=4*10/225=0.178m.
(From kx=mg)
Thus Amplitude= Max ext. - Mean ext.= 0.25-0.178 = 0.072m

If you really wanted to solve it with equations,
Spr. ext.= mg/k(ext. at mean) + A cos (wt)
v=- Aw sin(wt)
Since v=0, => sin(wt)=0, => wt=0 => t=0
Thus, 0.25=0.178+A
Thus A = 0.25-0.178 = 0.072m
Also generally, squaring and adding removes sin and cos IF THEY HAVE SAME COEFFICIENTS. You could also try to get it under one sine function using sin(a+b)
a sin(Z)+b cos(Z)= sqrt(a^2+b^2){sin(Z+tan-a/b)}

  • $\begingroup$ better to use dollar sign to show the equations in a pretty manner, now it is not much readable $\endgroup$ – jaromrax Feb 8 '17 at 12:49
  • $\begingroup$ @jaromrax, I wrote this answer when I didn't know LaTex existed. Will change it when I have the time. Thanks! $\endgroup$ – Dhruva Sambrani Jun 5 '18 at 11:56

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