An object undergoing simple harmonic motion has a maximum displacement of $6.2 \mathrm {~m}$ at $t = 0.00 \mathrm {~s}$ If the angular frequency of oscillation is $1.6 \mathrm { rad } / \mathrm { s } \text {, }$ what is the object's displacement when $t = 3.5 \mathrm {~s} ?$

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The correct answer is A.The equation of motion for an object undergoing simple harmonic motion is:$$x = A\cos(\omega t + \phi)$$where:* $x$ is the displacement of the object from its equilibrium position* $A$ is the amplitude of the motion* $\omega$ is the angular frequency of oscillation* $\phi$ is the phase angleIn this problem, we are given that the amplitude of the motion is $6.2 \mathrm {~m}$ and the angular frequency of oscillation is $1.6 \mathrm { rad } / \mathrm { s } \text {, }$. We are also given that the object is at its maximum displacement at $t = 0.00 \mathrm {~s}$. This means that the phase angle is $0$.To find the object's displacement when $t = 3.5 \mathrm {~s}$, we simply plug these values into the equation of motion:$$x = 6.2\cos(1.6 \times 3.5 + 0)$$$$x = 6.2\cos(5.6) = 4.8 \mathrm {~m}$$Therefore, the object's displacement when $t = 3.5 \mathrm {~s}$ is $4.8 \mathrm {~m}$.

An Object Undergoing Simple Harmonic Motion Has a Maximum Displacement 6.2 m6.2 \mathrm {~m}6.2 m

An Object Undergoing Simple Harmonic Motion Has a Maximum Displacement $6.2 \mathrm {~m}$