Question

For the simple harmonic motion equation d=5sin((pi/4)t), what is the period?. .

Answer 1

Answer:

8

Step-by-step explanation:

The given simple  harmonic motion equation is:

$d=5sin(\frac{\pi}{4})t$

Comparing the above equation with, $y=asinbx$, we have

a=5 and $b=\frac{\pi}{4}$

Now, period of the simple harmonic motion equation is given as=$\frac{2\pi}{|b|}$

=$\frac{2\pi}{|\frac{\pi}{4}|}$

=$8$

Thus, the period of the given simple harmonic motion equation $d=5sin(\frac{\pi}{4})t$ is 8.

Answer 2

Answer:

The period of given harmonic function is 8

Step-by-step explanation:

Given: The simple harmonic motion

$d=5\sin(\frac{\pi}{4}t)$

It is equation of simple harmonic motion.

We need to find the period of this function. It is sine function.

As we know the period of sine function is 2π

Period, sin x = 2π

If coefficient x is 1 then period is 2π

Period, sin (ax) = 2π/a

If coefficient of x is "a" then period is 2π divide by a

$\sin(ax)\rightarrow \sin(\frac{\pi}{4}t)$

$a\rightarrow \frac{\pi}{4}$

Period of given harmonic motion:

$\Rightarrow \dfrac{2\pi}{\pi/4}$

$\Rightarrow 8$

Hence, The period of given harmonic function is 8