What is the amplitude, period and the phase shift of #y = 2 sin (1/4 x)#?

2 Answers
Narad T.
May 26, 2018

The amplitude is #=2#. The period is #=8pi# and the phase shift is #=0#

Explanation:

We need

#sin(a+b)=sinacosb+sinbcosa#

The period of a periodic function is #T# iif

#f(t)=f(t+T)#

Here,

#f(x)=2sin(1/4x)#

Therefore,

#f(x+T)=2sin(1/4(x+T))#

where the period is #=T#

So,

#sin(1/4x)=sin(1/4(x+T))#

#sin(1/4x)=sin(1/4x+1/4T)#

#sin(1/4x)=sin(1/4x)cos(1/4T)+cos(1/4x)sin(1/4T)#

Then,

#{(cos(1/4T)=1),(sin(1/4T)=0):}#

#<=>#, #1/4T=2pi#

#<=>#, #T=8pi#

As

#-1<=sint<=1#

Therefore,

#-1<=sin(1/4x)<=1#

#-2<=2sin(1/4x)<=2#

The amplitude is #=2#

The phase shift is #=0# as when #x=0#

#y=0#

graph{2sin(1/4x) [-6.42, 44.9, -11.46, 14.2]}

Jim G.
May 26, 2018

#2,8pi,0#

Explanation:

#"the standard form of the sine function is"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=asin(bx+c)+d)color(white)(2/2)|)))#

#"amplitude "=|a|," period "=(2pi)/b#

#"phase shift "=-c/b" and vertical shift "=d#

#"here "a=2,b=1/4,c=d=0#

#"amplitude "=|2|=2," period "=(2pi)/(1/4)=8pi#

#"there is no phase shift"#

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