Vibrations of a spring Suppose an object of mass m is attached...

Question

Vibrations of a spring Suppose an object of mass m is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position $y equals 0$ when the mass hangs at rest. Suppose you push the mass to a position $y sub 0$ units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system), the position y of the mass after t seconds is $y = y_{0} \cos (t \sqrt{\frac{k}{m}})$ , where $k is greater than 0$ is a constant measuring the stiffness of the spring (the larger the value of $k$ , the stiffer the spring) and $y$ is positive in the upward direction.

Use equation (4) to answer the following questions.
c. How would the velocity be affected if the experiment were repeated with a spring having four times the stiffness ( $k$ is increased by a factor of $4$ )?

Use equation (4) to answer the following questions.

c. How would the velocity be affected if the experiment were repeated with a spring having four times the stiffness ( $k$ is increased by a factor of $4$ )?

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So, first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A bungee cord stretches and compresses as a person of mass M bounces up and down. The position Y of the person at time T is described by Y T is equal to A multiplied by cosine of T multiplied by the square root of K divided by M. Where A is the amplitude of oscillation, K is greater than 0, is the stiffness of the bungee cord, and YFT is positive when the person is above the equilibrium position. If the stiffness of the bungee cord is increased ninefold, K approaches 9K. How will this affect the velocity of the person during oscillation? Awesome. So that's our end goal. Our end goal is we're ultimately trying to figure out that when this stiffness of the bungee cord is increased ninefold, how will this affect the velocity of the person during this oscillation process? Awesome. So with that said, let's read off our multiple choice answers to see what our final answer might be. A is the velocity will decrease by a factor of 3. B is the velocity will increase by a factor of 3. See the velocity will increase by a factor of 9. And D is the velocity will remain the same. Awesome. So first off, in order to solve this problem, we need to differentiate YFT with respect to T in order for us to find VFT as a function of time. We also need to note that A, M, and K are all going to be constant values. We also need to recall and recognize that this particular problem requires the use of the change that we have to use the chain rule for differentiation, so we need to recall and use the chain rule. So, let us recall that the chain rule states. That D by DX. Of F of G of X. Is all equal to F of G of X multiplied by G of X. Awesome. So now we need to note that F of G of X is the outer function, so I'm going to put an O for outer function. And G of X is the inner function, so I'm gonna put a letter I for inner, just in case we've forgotten. So once again, you need to start thinking about the chain rule because we're gonna need to use the chain rule in order to perform this differentiation. Awesome. So with that said, we can now determine the outer function and its derivative. So the outer function in this case is going to be cosine of U, where in this case, we need to note that U is going to be equal to T multiplied by the square root of. Once again, this is the square root of K divided by M. And the derivative of the outer function with respect to you, so. So this is with respect to you, so this is gonna be once again the derivative of the outer function will be negative sign of you. Fantastic. So, therefore, you will be able to write that negative. Sign of U is equal to, once again, so negative sign of U is equal to negative sign of T multiplied by the square root of. K divided by M. Fantastic. So now we can determine officially the inner function and its derivative. So the inner function, as we should recall, and let's make a little note here that we're dealing with the inner, so we'll say I, so the inner function is T multiplied by the square root of K divided by M. And the derivative of the inner function, so now this is the derivative, I'll put a little letter D for derivative. So the derivative of the inner function with respect to T is just going to be simply the square root of K divided by M. So, therefore, without a doubt, we can write that DY by DT. is equal to V of T, which is equal to a multiplied by negatives of T multiplied by the square root of K divided by M. And then we need to multiply. This by the square root of K divided by M. Which is equal to, and we do a little bit of simplifying, we perform the calculation, 9, sorry, negative, so it's going to be negative A. Multiplied by. OK, so, so it's gonna be negative A multiplied by the square rooted K divided by M multiplied by the sign of T multiplied by the square root of K divided by M. Awesome. So now, our next step is we need to apply the change in stiffness. So now we're considering When K approaches 9K. So, this once again we're applying the change in stiffness. So our Vnu of T is going to be equal to negative A, so it's going to be equal to negative A multiplied by the square root of 9k divided by M multiplied by the sine of T multiplied by the square root of 9k divided by M. Fantastic. So that will mean that our V new of T, so this is our new value, so our Vnu of T when we perform the calculation is going to be equal to -3 multiplied by A, multiplied by the square root of K. Divided by M multiplied by the sign of 3 multiplied by T, multiplied by the square root of K divided by M. So, therefore, as we can see, as we should recall a note, when we increase the stiffness of this bungee cord ninefold, it will result in tripling the magnitude of the velocity due to the factor of 3 in the amplitude. So therefore we also need to note and recall that this will also triple or triple the frequency of the oscillation due to the factor of 3 inside of the sine function. So therefore, when we look at our multiple choice answers, the correct answer has to be. The letter B, the velocity will increase by a factor of 3, and this is because when we were solving below, we had it 3 multiplied by the amplitude. And that's it. We've officially solved for this problem. Hooray. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Key Concepts

Simple Harmonic Motion (SHM)

Spring Constant (k)

Velocity in SHM

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