Question

A horizontal spring is lying on a frictionless surface. One end of the spring is attaches to a wall while the other end is connected to a movable object. The spring and object are compressed by 0.080 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 12.1 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.041 m relative to its unstrained length

Answer 1

-0.831 m/s

Step-by-step explanation:

With amplitude A = 0.08 m and angular frequency ω = 12.1 rad/s, we have the following simple harmonic equations of motion:

Distance: x(t) = Acos(ωt) = 0.08cos(12.1t)

Speed: v(t) = -Aωsin(ωt) = -0.08*12.1sin(12.1t) = -0.968sin(12.1t)

At distance = 0.041, we can solve for 12.1t

x(t) = 0.041

0.08cos(12.1t) = 0.041

cos(12.1t) = 0.5125

12.1t = 1.033

t = 0.085s

Use it to plug into the v(t) equation

v(0.085) = -0.968sin(12.1*0.085) = -0.968*sin(1.033) = -0.968*0.859 = -0.831 m/s