what is the amplitude and the period of these problems?

Question

y=1/2 sin(3x)
y=3 cos(1/2x)

Answer 1

For the general equations y=asin(bx) and y=acos(bx), the amplitude is a and the period is 2pi/b. In these cases, the amplitudes are 1/2 and 3, and the periods are 2/3*pi and 4*pi.

Answer 2

The equation you work with is

y=asinx(b-c)+d
or
y=acosx(b-c)+d

a is amplitude, b is your period, c is your x axis shift, and d is your vertical shift.

The amplitude is the |a|, so you will never have a negative amplitude. Remember there is always an amplitude too, so there is no such thing as 0 amp. So for your first problem, your amplitude would be 1/2 and for your second problem, it would be 3.

Now period is determined by 2π/|b|. Since your first problems b is 3, your answer would be 2π/3 or 120. Your second problem has b as 1/2. Your problem would be 2π/1/2, or 2π *2. (multiply by the reciprocal.) So your period would be 4π or 720.

Hope This Helped.

Answer 3

the amplitute is obtained from the number preceding the function. the period is obtained by divding 2pi by the number preceding the x.

1) y = 1/2 sin(3x),
ampplitude = 1/2
period = 2pi/3 which also equals 120 degrees

2)y = 3cos(1/2x)
amplitude = 3
period = 2pi/1/2 = 4pi which also equals 720 degrees.

Answer 4

1/2 sin(3x), the period is 3/2pi
3 cos (1/2 x), the period is 1/4pi

The peak to peak amplutude of 1/2 sin(3x) is -1/2 to + 1/2 which is 1

The peak to peak amplitude of 3 cos(1/2 x) is -3 to +3 which is 6

Answer 5

1) A = 1/2 and 3= 2pi/T so T = 2pi/3
2) A= 3 1/2 = 2pi/T so T =4pi