How to prove the maximum displacement of a spring motion function

Key takeaway: For sinusoidal motion, the key quantity is the amplitude. You can read it directly from the coefficients, or prove it carefully by rewriting the expression as a single cosine or sine term. Both approaches point to the same maximum displacement.

We are given

$f(t)=\frac13\cos 12t-\frac14\sin 12t.$

We want to prove that the maximum displacement is $\frac{5}{12}$ inches.

1) Recognize the sine-cosine form

A function of the form

$A\cos(\omega t)+B\sin(\omega t)$

has amplitude

$\sqrt{A^2+B^2}.$

Here,

$A=\frac13, \qquad B=-\frac14.$

So the amplitude is

=\sqrt{\frac19+\frac1{16}}.$$ Find a common denominator: $$\frac19+\frac1{16}=\frac{16+9}{144}=\frac{25}{144}.$$ Thus $$\sqrt{\frac{25}{144}}=\frac{5}{12}.$$ ### 2) Why this is the maximum displacement We can also justify it directly using the identity $$\left|\frac13\cos 12t-\frac14\sin 12t\right|\le \sqrt{\left(\frac13\right)^2+\left(\frac14\right)^2}=\frac{5}{12}.$$ This inequality is a standard consequence of the Cauchy–Schwarz inequality or the amplitude form of a sinusoid. Therefore the maximum possible value of $|f(t)|$ is $\frac{5}{12}$, so the maximum displacement is $$\boxed{\frac{5}{12}\text{ inches}}.$$ --- **Pitfalls the pros know** 👇 A frequent mistake is to take the largest coefficient, $\frac13$, and assume that is the amplitude. It isn’t. In a sum of sine and cosine terms, the coefficients combine by the square root of the sum of squares. Another small trap is confusing maximum value with maximum displacement: displacement usually means the maximum of $|f(t)|$, not just the largest positive value. **What if the problem changes?** If the function were written as $a\cos 12t+b\sin 12t$, the amplitude would always be $\sqrt{a^2+b^2}$. If a phase shift were added, like $\frac{5}{12}\cos(12t-\phi)$, the maximum displacement would stay the same; only the timing of the peaks would change. `Tags`: amplitude, simple harmonic motion, trigonometric identity