Suppose that an object undergoes simple harmonic motion, and its displacement has an amplitude A = 15.0 cm and a frequency f = 1
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Answer:
a.) magnitude __49.7__ unit(s)
b.) direction __123.6°_ counterclockwise from the +x axis
Explanation:
Let Vector is v
x-component of Vector v = x = -27.5 units (minus sign indicate that x-component is along the minus x-axis )
y-component of Vector v = y = 41.4 units
Magnitude of v = ?
Direction of v = ?
To find the magnitude of the vector
v =
v =
v = 49.7 units
To find direction
θ = tan⁻¹(y/x)
θ = tan⁻¹(41.4/-27.5)
θ = -56.4°
This Angle is in the clockwise direction with respect to -x axis.
We need to find Angle counterclockwise from the +x axis.
So,
θ = 180° - 56.4°
θ = 123.6°
The given vector is in 2nd quadrant
Use one of the equations of accelerated motion; V2=V1 + at ...see attached
