# maximum displacement

#### bobby2017

A simply supported beam is subjected to two vibrations along its length, emanating from two machines at opposite ends of the beam. The displacement caused by the vibrations can be modelled by the following equations.
x1=5sin(100π+π/2)
x2=6.5sin(100π-π/3)
when both machines are switched on, how long does it take for each machine to produce its maximum displacement?

#### skeeter

Math Team
x1=5sin(100π+π/2)
x2=6.5sin(100π-π/3)
shouldn't those two equations be functions of time, $t$ ?

#### skeeter

Math Team
For each machine individually ...

$x_1$ starts out as with a maximum positive displacement of 5 at $t=0$. It achieves a max negative displacement of -5 at at $t = \dfrac{1}{100}$ sec (half the period) and back to a max positive displacement of 5 at $t=\dfrac{1}{50}$ sec. This is due to the fact that $x_1$ has the simplified equation of $x_1 = 5\cos(100\pi t)$

$x_2$ achieves a max positive displacement of 6.5 at $t = \dfrac{1}{120}$ sec

Normally, a question like this asks to find the time and value of the max displacement from equilibrium of the superposition, i.e. the sum, of both waves $x_1+x_2$, which would require knowledge of wave addition techniques. As written, question (b) is a bit vague on this point ...

#### bobby2017

I've found someone asking near enough the same question, just with a few number changes.
I just don't understand that math, to be honest.

#### skeeter

Math Team
The Socratic post solution uses a method of calculus to determine the maximum value, which really isn't necessary.

You do understand the maximum value of the sine function is $\sin\left(\dfrac{\pi}{2} + 2k\pi \right) = 1$, where $k$ is an integer value, correct?

So, $x_1 = 5\sin\left(100\pi t + \dfrac{\pi}{2} \right)$ has a max value of 5 when $100\pi t + \dfrac{\pi}{2} = \dfrac{\pi}{2} + 2k\pi$

That equation is true when $100\pi t = 2k \pi \implies t = \dfrac{k}{50}$ ... the first value of time > 0 when that occurs would be $t = \dfrac{1}{50}$

Note the period of $x_1$ is $T = \dfrac{2\pi}{100\pi} = \dfrac{1}{50}$, which is at the end of the wave's first period.

$x_2 = 6.5\sin\left(100\pi t - \dfrac{\pi}{3} \right)$ has a max value of 6.5 when $100\pi t - \dfrac{\pi}{3} = \dfrac{\pi}{2} + 2k\pi$

see what you can do yourself from here ...