Question

pleaseee helpp !!!!what is the period and amplitude of the function for y=3 sin 4x

Answer 1

amplitude is the absolute value of the number sitting out front

Answer 2

period of
\[\sin(bx)\] is
\[\huge \frac{2\pi}{b}\]

Answer 3

have you tried drawing it? start with y = sin(x), then draw y = 3 sin(x), then the harder part draw y = 3 sin(4x).

by draw, a simple doodle will suffice, and i am happy to help, but doing that really fires the synapses!

Answer 4

$$
y=A\sin(\omega t)\\
$$
Where \(\omega\) is the angular frequency: \(\omega = {2\pi\over T}\), \(T\) is the period and \(A\) is the Amplitude.
So
$$
\omega = b\\
\implies {2\pi\over T}=b\\
\implies T={2\pi\over b}\text{,which is the period.}\\
$$
It should be clear from this what the value of \(A\) is to make the following equal to your equation:
$$
y=A\sin \left ({2\pi \over T}\cdot x \right)={\color{red}{A}}\sin bx={\color{red}{3}}\sin bx
$$

Answer 5

y=asin(b(x-c))+d
a is your amplitude but it is the absolute vaue of a
so
example y=-2sin(4x) the amplitude is 2
your period is (2pi)/b in this case b is 4 just sub
(2pi)/4 which is pi/2 can you apply this to your problem?

Answer 6

d is the midline and c is your phase shift

Answer 7

@pham96 post this agin here, or on the physics forum, if this has been unhelpful. cheers!