Question

The above stress can be also approximated by a sinusoidal function (i.e. sine wave) with an average 3,811 kPa, amplitude 430 kPa, and a frequency of 6 Hz. At a second flight condition, the stress had an additional sinusoidal component (i.e., in addition to the first sine wave above). This second component is caused by half the amount of forces (relative to the first condition), has three times the frequency, and 90 degrees out of phase (leading). Determine the average of the second sinusoidal component in kPa. (Provide the answer using 2 decimal places).

Answer 1

306.08 kPa

Step-by-step explanation:

Given data

Average stress of first wave = 3811 kPa

amplitude of first wave ( A1 ) = 430 kPa

Frequency of first wave = 6 Hz

Determine the average stress of the second sinusoidal component in kPa

Amplitude of the additional wave (A2add) =
`(A1)/(2)` = 430 / 2 = 215 kPa

next we will determine the resultant amplitude of the second wave

A2 =
`\sqrt{A^(2) _(1) } + A^(2) _(2add) + 2A_(1)A_(2add) Cos\alpha`

=
`√(430^2+ 215^2 + 2(430*215) * cos 90)`

=
`√(184900 + 46225 )`

=
`√(231155)` = 480.79

hence the average stress of the second sinusoidal component

=
`(2A_(2) )/(\pi )`

=
`(2 * 480.79 )/(\pi )` = 306.08 kPa