`y = 2sin(5x)` Find the period and the amplitude

Asked on by nick-teal

Textbook Question

Chapter 4, 4.5 - Problem 5 - Precalculus (3rd Edition, Ron Larson).
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steveschoen's profile pic

steveschoen | College Teacher | (Level 1) Associate Educator

Posted on

This would be an example of what's called a sinusoidal function.  The general form for that is:

y = A*sin(Bt-C)+D

Where A --> amplitude

B --> period = 2*pi/B

C --> phase shift = C/B

D --> vertical shift

So, simply matching the numbers up, the amplitude would be 2.  And, the period would be 2*pi/5 = 0.4*pi.

t-rashmi's profile pic

t-rashmi | College Teacher | eNotes Newbie

Posted on

As this question has been posted in the Maths section, I am assuming that it is not related to Simple Harmonic Motion.

If a function f(x) has a period = t, then the function n*f(kx) has a period = t/k

We know that the function f(x) = sin(x) has a period = 2*pi. Thus the function f(x) = 2sin(5x) will have a period = (2/5)*pi.

The maximum value of sin(5x) is 1 and the mean value is 0. Thus the amplitude of y=2sin(5x) is equal to 2

Refer to the graph attached for better understanding.

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