Astronauts in space 'weigh' themselves by oscillating on a spring. Suppose the position of an oscillating 75 kg astronaut is given by $x = (0.30 m) \sin ((π rad/s) \times t)$ , where t is in s. What force does the spring exert on the astronaut at (a) t = 1.0 s and (b) 1.5 s? Note that the angle of the sine function is in radians.

Question

Accepted Answer

Hey, everyone. So this problem is working with spring forces. Let's see what it's asking us, Jimmy and his friend make a contraption with a horizontal mass spring system. In the first trial, they attach a 68 kg mass to the spring and set it to oscillate. Assume that the oscillating objects position is given by the equation X equals 0.240 m multiplied by the cosine of pi radiance per second, multiplied by time At time equals two seconds. How much force does this spring exert on the object? Our multiple choice answers here are a, our net force is equal to negative 16.32 pi squared B net force is equal to negative 14.42 pi squared C net force is equal to negative 21.68 pi squared or D net force is equal to negative 35.41 pi squared. OK. So the first thing we're going to do here is recognize that our acceleration is given by that derivative of velocity over the derivative of time or D V divided by D T. And the derivative of velocity in turn, if we take an, if we take the second derivative, that is Our 2nd derivative of X or position divided by D T squared. So just recalling the relationship here between acceleration, velocity and position, so that is key because Newton's second law tells us that the force is equal to mass multiplied by the acceleration. We aren't given acceleration in this problem, but we are given a position function. So using this relationship between acceleration and position, we can determine the acceleration function. And then in turn solve for the force. So that is exactly what we are going to do here. So we have velocity is equal to D X over D T. So when we take the derivative of this function given to us in the problem, we are left with negative 0.24 m multiplied by hi multiplied by the sign of pi radiance per second multiplied by time. And then acceleration is the velocity is the derivative of the velocity over time. So we're gonna take this equation here and derive it again with respect to time. And that gives us negative .2, 4 m multiplied by pi squared multiplied by the cosine of pi radiance per second multiplied by time. And so now we have our acceleration equation and we can plug in this, we can plug in our time of two seconds to find the acceleration at two seconds. So a the acceleration of two seconds is equal to negative is equal to negative . m multiplied by pi squared multiplied by the cosine of pie radiance per second multiplied by two seconds. And we plug that into our calculator and we get an acceleration of negative .24 m multiplied by pi squared Because the cosine of two pi is one. So back to our Newton second law equation, then now we have acceleration we can solve for the force force equals mass multiplied by the acceleration, the mass. And the problem was given as 68 kg Multiplied by negative .24 m multiplied by pi squared. And that gives us negative 16.32 pi squared. And so that is the correct answer for this problem. And that aligns with answer choice A so that's all we have for this one. We'll see you in the next video.

Verified video answer for a similar problem:

Key Concepts

Simple Harmonic Motion

Hooke's Law

Force Calculation in Oscillatory Motion