Verified Answer :

boffeemadrid

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.

isyllus

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

Related Questions