Question

Find a model for simple harmonic motion if the position at t=0 is 5 centimeters, the amplitude is 5 centimeters, and the period is 4 seconds.
A) d=5cos(4t)
B) d=5sin(pi/2t)
C) d=4sin(5t)
D) d=5cos(pi/2t)

Answer 1

Answer:

$\boxed{\boxed{D.\ d=5\cos \left(\dfrac{\pi}{2}t\right)}}$

Step-by-step explanation:

As at t=0 the amplitude is 5 cm, so the simple harmonic motion will be in cosine form. As in case of sine function, the value of function at x=0 is 0.

$\sin 0=0$

We know that, in the function

$y=a\cdot \cos b(x+c)+d$

1. period is $\dfrac{2\pi}{b}$,
2. horizontal shift or phase shift is c,
3. amplitude is a,
4. vertical shift is d.

As in this case the amplitude is given to be 5, hence $a=5$

Also the period is given as 4, so

$\Rightarrow \dfrac{2\pi}{b}=4$

$\Rightarrow b=\dfrac{2\pi}{4}=\dfrac{\pi}{2}$

As nothing is given about the horizontal and vertical shift, so putting $c=0,d=0$, the function becomes,

$y=5\cdot \cos \dfrac{\pi}{2}(x+0)+0$

or $y=5\cos \left(\dfrac{\pi}{2}x\right)$

As in x axis time is taken as t and in y axis distance is taken as d, so the function becomes,

$d=5\cos \left(\dfrac{\pi}{2}t\right)$

Answer 2

It would be b bc if you do all of ur steps