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MCAT Physical : General Principles and Properties

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Example Questions

Example Question #62 : Waves

At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of $v = 331 \frac{m}{s} + 0.6\frac{m}{s}(T)$ $\displaystyle v = 331 \frac{m}{s} + 0.6\frac{m}{s}(T)$, where T is the temperature in °C.

How long would it take for a 30Hz beat to reach an audience member 100m away when the ambient temperature is 21ºC?

Possible Answers:

0.31s

0.29s

0.91s

0.11s

Correct answer:

0.29s

Explanation:

The question asks us to determine how long it will take for a wave beat to reach an audience member at 100m away; thus, we need to calculate the velocity of the wave to determine the time.

We know from kinematics that $v = \frac{d}{t}$ $\displaystyle v = \frac{d}{t}$. This can be rearranged to solve for t: $t = \frac{d}{v}$ $\displaystyle t = \frac{d}{v}$.

$v = 331 \frac{m}{s} + 0.6\frac{m}{s}(T)$ $\displaystyle v = 331 \frac{m}{s} + 0.6\frac{m}{s}(T)$

$v = 331 \frac{m}{s} + 0.6\frac{m}{s}(21) = 343.6\frac{m}{s}$ $\displaystyle v = 331 \frac{m}{s} + 0.6\frac{m}{s}(21) = 343.6\frac{m}{s}$

$t = \frac{d}{v}=\frac{100m}{343.6\frac{m}{s}}=0.29s$ $\displaystyle t = \frac{d}{v}=\frac{100m}{343.6\frac{m}{s}}=0.29s$

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Example Question #1 : Velocity And Waves

A wave produced on a string travels with a velocity of $500\frac{m}{s}$ $\displaystyle 500\frac{m}{s}$. If the tension on the string is increased by a factor of four, at what speed does the wave travel?

Possible Answers:

$125\frac{m}{s}$ $\displaystyle 125\frac{m}{s}$

$2000\frac{m}{s}$ $\displaystyle 2000\frac{m}{s}$

$500\frac{m}{s}$ $\displaystyle 500\frac{m}{s}$

$1000\frac{m}{s}$ $\displaystyle 1000\frac{m}{s}$

Correct answer:

$1000\frac{m}{s}$ $\displaystyle 1000\frac{m}{s}$

Explanation:

The velocity of a wave can be obtained with the formula $v=\sqrt{\frac{T}{\mu }}$ $\displaystyle v=\sqrt{\frac{T}{\mu }}$, where $\displaystyle T$ is the tension in the string and $\mu$ $\displaystyle \mu$ is the mass per unit length of the string. Since the tension is quadrupled, the velocity will be doubled.

Let's assume that a string with tension $\displaystyle T$ and a mass per unit length $\mu$ $\displaystyle \mu$ produces a wave with velocity $500\frac{m}{s}$ $\displaystyle 500\frac{m}{s}$.

$v_1=500 \frac{m}{s} =\sqrt{\frac{T}{\mu}}$ $\displaystyle v_1=500 \frac{m}{s} =\sqrt{\frac{T}{\mu}}$

If we increase the tension by a factor of four, we will get the below expression.

$v_2=\sqrt{\frac{4T}{\mu}}= 2 * \sqrt{\frac{T}{\mu}}$ $\displaystyle v_2=\sqrt{\frac{4T}{\mu}}= 2 * \sqrt{\frac{T}{\mu}}$

We can see that $\displaystyle v_2=2*v_1$, and we know that $v_1=500\frac{m}{s}$ $\displaystyle v_1=500\frac{m}{s}$.

$v_1=\sqrt{\frac{T}{\mu}}=500\frac{m}{s}$ $\displaystyle v_1=\sqrt{\frac{T}{\mu}}=500\frac{m}{s}$

$v_2=2*\sqrt{\frac{T}{\mu}}=2*v_1=2*(500\frac{m}{s})$ $\displaystyle v_2=2*\sqrt{\frac{T}{\mu}}=2*v_1=2*(500\frac{m}{s})$

$v_2=1000\frac{m}{s}$ $\displaystyle v_2=1000\frac{m}{s}$

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Example Question #31 : Ap Physics 2

Through which of the following would you expect a photon to travel fastest?

Possible Answers:

Glass

Air

Water

Vacuum

Correct answer:

Vacuum

Explanation:

A photon will travel fastest through a vacuum. Photons are generally massless and can be thought of as a light wave, which travels fastest in a vacuum and slowest through a metal or solid. This can be visualized using the concept of the index of refraction, which describes the speed of light through air compared to the speed through other mediums. A vacuum will be the least dense and cause the least hindrance to a photon as it travels, thus giving it the lowest index of refraction and allowing the fastest speed of light.

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Example Question #66 : Waves

You are out snorkling off the coast of an exotic island when a pod of whales comes swimming by. The pod is 100m away. If they emit sounds underwater with an average frequency of 2200Hz and there are 500 complete sound waves between you and the pod, how fast is the speed of sound in the water?

$v_{s_{air}}=340 \frac{m}{s}$ $\displaystyle v_{s_{air}}=340 \frac{m}{s}$

Possible Answers:

$300 \frac{m}{s}$ $\displaystyle 300 \frac{m}{s}$

$440 \frac{m}{s}$ $\displaystyle 440 \frac{m}{s}$

$380 \frac{m}{s}$ $\displaystyle 380 \frac{m}{s}$

$510 \frac{m}{s}$ $\displaystyle 510 \frac{m}{s}$

$620\frac{m}{s}$ $\displaystyle 620\frac{m}{s}$

Correct answer:

$440 \frac{m}{s}$ $\displaystyle 440 \frac{m}{s}$

Explanation:

If there are 500 waves over a distance of 100 meters, we can say that the wavelength is:

$\lambda = \frac{100m}{500} = 0.2 m$ $\displaystyle \lambda = \frac{100m}{500} = 0.2 m$

Now we can use the formula for the speed of waves:

$v = \lambda f = (0.2m)(2200s^{-1}) = 440 \frac{m}{s}$ $\displaystyle v = \lambda f = (0.2m)(2200s^{-1}) = 440 \frac{m}{s}$

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Example Question #21 : General Principles And Properties

Which statement is not true for all waves?

Possible Answers:

$v=f\lambda$ $\displaystyle v=f\lambda$

Their speed depends on the temperature of the transmitting medium

They can produce interference

They can be diffracted

Wavelength is the distance between corresponding points on successive cycles

Correct answer:

Their speed depends on the temperature of the transmitting medium

Explanation:

The speed of sound is dependent on the temperature of the transmitting medium. The speed of light is not.

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MCAT Physical : General Principles and Properties

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #62 : Waves

Example Question #1 : Velocity And Waves

Example Question #31 : Ap Physics 2

Example Question #66 : Waves

Example Question #21 : General Principles And Properties

All MCAT Physical Resources

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