SHM question and Current question.

[al_201314]

Hi guys,

I'm stuck at this 2 questions for a while.

1. A mass suspended from a vertical spring is displaced vertically downwards by a distance A and released. It executes simple harmonic motion about its equilibrium position O. Given that time taken for the mass to move through a distance of 0.5A from the point of release is t, what is the period of the oscillation of the mass?

I drew the cosine curve and at 0.5A, the corresponding time is a unknown, but it is >T/8 where T is the period. How do I carry on from here?


2. Over a period of 6 seconds, a current is uniformly reduced from 150mA to 30mA. WHat is the charge that flows during this time?

The worked solutions is (150-30)6 = 720mC.
720-150-30 = 540mC which is the answer.

Why is this so? I would have thought that 720 is the reduction in charge? And why the subtraction of 150 and 30?

Thanks for the help in advance.

 

[quasar987]

1. A mass suspended from a vertical spring is displaced vertically downwards by a distance A and released. It executes simple harmonic motion about its equilibrium position O. Given that time taken for the mass to move through a distance of 0.5A from the point of release is t, what is the period of the oscillation of the mass?

I drew the cosine curve and at 0.5A, the corresponding time is a unknown, but it is >T/8 where T is the period. How do I carry on from here?

You have to draw from your knowledge of the original sine function, for which you know that the period is $2\pi$. If you can find at what angle 'a' the function $x(t')=-A\sin (t')$ is worth -0.5A, then you can see what ratio this angle makes with the period of oscillation $2\pi$.

And since the motion of the mass is also a sinusoidal one with the only difference that the period is $T=\sqrt{m/k}2\pi$ instead of $2\pi$, t is in the same ratio with T as 'a' is with $2\pi$.

 

[quasar987]

2. Over a period of 6 seconds, a current is uniformly reduced from 150mA to 30mA. WHat is the charge that flows during this time?

The worked solutions is (150-30)6 = 720mC.
720-150-30 = 540mC which is the answer.

I can't answer your question directly, but I can show you a method of finding the solution that doesn't involve banging your head on the wall.

We know that the current is reduced uniformly. This means that the graph of I vs t is a line of negative slope. We can thus write

$$I(t)=-kt + I_0$$

We are told that I(0)=150mA and I(6)=30mA. Plugging this into the equation, we find that $I_0=150$ mA and $k=20$mA/s:

$$I(t)=-20t + 150$$

We also know that the current is the amount of charge that flows per unit time: $I=dq/dt$. A convenient picture to have is to regard the wire in which the current flows as a hose, but instead of water coming out of the hose, it is charges. For instance, if the current in the wire is of 1A, it means that 1 coulomb of charge is oosing out of the hose each second.

Setting $I=dq/dt$ in our equation, we then have

$$\frac{dq}{dt}=-20t + 150$$

Since both sides of the equation are equal, the integral of the right side equals the integral of the left side:

$$\int \frac{dq}{dt}dt=\int (-20t + 150)dt$$

Using the fondamental thm of calculus on both integrals, we get

$$q(t)=-10t^2+150t+q_0$$

What does q(t) represent? Following our hose analogy, it is the quantity of charge (in mC!) that has flown out of the hose after a time t. We want to know, then, the amount of charge that has flown out between time t=0 and t=6. It is

$$\Delta q = q(6)-q(0)=10(36)+150(6)+q_0-q_0= 540 \mbox{mC}$$

 

[mezarashi]

For part 1, I think you are on the right track. Consider your cosine curve again. Say that the peak value is A (on the y-axis). You are given the time T when the curve is at 1/2 A.

y = f(t)
A = cos (t=0)
1/2 A = cos (2piT/T0)

You can solve for T0, the period, in terms of A and the given T.

For part 2, you are incorrect because you are not using calculus. Remember, the flow of current is changing. Fortunately, the change is uniform. What is the average current flowing during this 6 second period? You've arrived at 540 incorrectly. See if you can make sense out of:

(150+30)/2 * 6 = 540

Good luck ^^

 

[al_201314]

Thanks for the replies guys. Part 1's going good for me except for the portion after integrating, why is there a +qnot in the equation?

For part 2, I still cannot fully get it. Here's what I have after reading the posts and some thinking through.

0.5A = Acos(2pit/T). The A cancels out on both sides and taking the inverse cosine function on the LHS, I resolved time t = 1.047, since T = 2pi which cancels out with the top.

Then, similarly, for a full period, A = Acos(2pit/T) which I have t = 1.571. The ratio of t/T = .0667. How do I carry on?

Thanks agaiN!

 

[mezarashi]

Thanks for the replies guys. Part 1's going good for me except for the portion after integrating, why is there a +qnot in the equation?

For part 2, I still cannot fully get it. Here's what I have after reading the posts and some thinking through.

0.5A = Acos(2pit/T). The A cancels out on both sides and taking the inverse cosine function on the LHS, I resolved time t = 1.047, since T = 2pi which cancels out with the top.

Then, similarly, for a full period, A = Acos(2pit/T) which I have t = 1.571. The ratio of t/T = .0667. How do I carry on?

Thanks agaiN!

The T that I used represented the period. I think that is what you are suppose to find. Why are you assuming it as 2pi?

For part2: How about we turn this into a mechanics question. At t=0, you are driving at 150km/hr. In 6 hrs, you slow down to 30km/hr uniformly. This means that at t=1 or 2 or 3 or 4 or 5, your speed is somewhere between 150 and 30. Try drawing a uniform line between these two speeds (it will look like a right triangle). Now, can you find the average speed from t=0 to t=6?

If I travel at an average speed of 90km/hr for 6hrs, how far would I go? Now similarly, change the units to 150mA or 150mC/s, and approach the problem similarly.

 

[quasar987]

(The q0 is the constant of integration. Physically, it represents the amount of charge that oozed out of the hose prior to t=0. Since we could always assume that the current was turned on instantaneously to 150mA at t=0, we could take q0=0. Or not, since it cancels out anyway when we calculate $\Delta q$)