Question

The above stress can be approximated by a sinusoidal function (ie, sine wave) with an average 2,388 kPa, amplitude 293 kPa, and a frequency of 6 Hz. At a second flight condition, the stress had an additional sinusoidal component (ie, in addition to the first sine wave above) that has three times the above frequency, half of the magnitudes, and 90 degrees out of phase (leading). Determine the amplitude of the second sinusoidal component in kPa. Provide the answer using 2 decir placed.

Answer 1

327.58 kPa

Step-by-step explanation:

Given data; 1st wave

Amplitude ( A1 ) = 293 Kpa

frequency = 6Hz

Average stress = 2388kPa

phase angle = 90° ( leading )

Determine the amplitude of the second sinusoidal component in kPa

Amplitude of additional wave ( A2 ) = 293 / 2 = 146.5 kPa

Amplitude of first wave ( A 1 ) = 293 kPa

hence the amplitude of the second sinusoidal component

A' =
`\sqrt{} A^(2) _(1) } + A^2_(2) + 2A_(1) A_(2) *cos 90`

=
`\sqrt293^2 + 146.5^2 + 2(293*146.5) *( -0.45)`

=
`√(85849+ 21462.25)`

= 327.58 kPa