## Session 88

As always, Bryan from Next Step Test Prep is joining us today as we start off a sequence of episodes – all about physics!

Physics is one of the hardest subjects on the MCAT. Over the course of 4 episodes, we’re going to dive into high-yield physics topics to help you score higher!

What makes so hard for students, as Bryan explains, is that metaphorical “hard science” meaning a lot of numbers, fact, or truth to it. And for MCAT students, it gets hard because of the math primarily. Some students are not really into math, and add pressure to that where you get no calculator on test day.

Plus, a lot of students to take Physics really early so when they roll around in their MCAT by Junior or Senior year or even after college, it’s been a long time since they’ve taken Physics.

Lastly, Physics is like the “orphan” science that lives over on the side that it’s not often obviously directly plugged into Chemistry or Organic Chemistry, Biochem, and Biology. And you can create this whole matrix of all your other classes where the information relate to each other. And on the side, Physics asks questions about, say, shooting cannonballs in the air.

### [03:30] Pendulum and Oscillation

Question 6: By what factor does the period of oscillation of a pendulum change when its length is extended from 1 cm to 9 cm?

- (A) Original period is multiplied by 0.33.
- (B) It’s original period is multiplied by 2.
- (C) It’s multiplied by 3.
- (D) It’s multiplied by 9.

*Bryan’s Insights:*

If you have a pendulum that’s a lot longer, it would swing back and forth slower. And remember that period is the time it takes to oscillate all the way out and then back again. So if it has to go slower, then the period would have to increase as well.

Thinking about how equations are arranged, if you took one of the variables and give it a 9x times 9, B does not seem to fit. And even if you can’t remember the equation, you’d probably guess it to be C or D. In this case, the correct answer is C because of the square root relationship. The period is proportional to the square root of the length.

So when you multiply length times 9, period goes up by 3. The exact equation here is:

**T** = 2π**√**(**L**/**g**)

Where: T = period, L= length, g = gravity

### [06:44] Periodic Motion

Question: A spring system shown below (diagram of a mass at the end of a spring resting horizontally on a table), with Mass (m), representing the mass attached to the spring, (k) representing the spring constant, and (x) denoting the distance, stretched or compressed from the spring’s equilibrium position. What expression gives the minimum value for the magnitude of velocity of the mass after the system is stretched and released?

- (A) 0
- (B) 1.2 kx²
- (C) x(
**√**(k/m) - (D) x²k/m

*Bryan’s Insights:*

This can still show up on the MCAT, like if they say, bone is slightly compressible and can be modeled as a spring and so on.

What you want to know here is that when you oscillate back and forth whether you’re a spring vibrating back and forth like the car shocks bouncing up and down. Or rather a pendulum swinging back and forth. In any kind of periodic motion like this, you want to be aware of what happens at the extremes like when the pendulum swings all the way up, or when the spring compresses all the way down, or vice versa.

When happens in these periodic motion systems at the extremes, a pendulum for instance, when the mass swings all the way up in a pendulum, velocity goes zero and stops moving. Then it starts swinging back down. This is the same thing here.

This may look like a crazy algebra question, when in fact, you only had to know one concept about how periodic motion works. This means that at the extremes, you stop for a second, turn around, and start going back. So the answer here is (A), with no algebra at all.

Let’s go over the other answer choices above to make sure you recognize them. (B) is the measure of the energy stored in a spring. Students may see this and and tend to pick this but this is not the minimum possible velocity.

### [11:08] Stretching the Spring

Question 10: A certain elastic coil has a spring constant of 100 Newtons per meter. How much must this spring be stretched in order to store the same amount of energy held by a 50-kg mass at rest, 1m above the surface of the earth?

- (A) 10 cm
- (B) 31 cm
- (C) 3.1m
- (D) 10 m

*Bryan’s Insights:*

Bryan explains this being a two-step problem. There’s an equation for energy which is E=mgh.

So 50 x 10 (for gravity) x 1 = 500 J of energy.

Then you can re-read the question knowing that what you need is 500 J of energy. So the k is the spring constant of 100 Newtons.

You have to remember that the energy stored in the spring is 1/2 kx². So you’ve got 500=1/2 kx². So right about 3 is your answer (C).

Additionally, you’re going to be working on SI units on the MCAT. So the base unit for length is meter. The big takeaway actually is to know all of your equations. But certainly when you’re doing Physics, this is going to be a big part of your problem solving.

### [16:30] Do Practice Full Lengths

The AAMC has full lengths for sale. They make the MCAT so their test writers also have a practice exams which you can buy from them directly. You can buy three practice exams plus a sample task that’s not scored.There are conversion tables you will find online but don’t trust those. For a full view of resources that may help you.

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