Answer to Question #346683 in Physics for blueming

Question #346683

The velocity of a particle undergoing simple harmonic motion is described by,

𝑣(𝑡) = 2.25m/s cos(0.555 rad/s∙ 𝑡 + 0.450 rad) What is the particle’s maximum acceleration? 


1
Expert's answer
2022-06-06T08:55:34-0400
"\ud835\udc63(\ud835\udc61) = 2.25 {\\:\\rm m\/s} \\cos\\left((0.555 {\\:\\rm rad\/s}) \ud835\udc61 + 0.450 {\\:\\rm rad}\\right)"

The acceleration

"a(t)=\ud835\udc63'(\ud835\udc61) = 2.25 {\\:\\rm m\/s} *(0.555 {\\:\\rm rad\/s}) \\\\\n*\\sin\\left((0.555 {\\:\\rm rad\/s}) \ud835\udc61 + 0.450 {\\:\\rm rad}\\right)"

"a_{\\max}= 2.25 {\\:\\rm m\/s} *(0.555 {\\:\\rm rad\/s})=1.25\\:\\rm m\/s^2"


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