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A 3.49-rad/s ( \(33 \frac{1}{3}\) rpm) record has a \(5.00-\mathrm{kHz}\) tone cut in the groove. If the groove is located \(0.100 \mathrm{m}\) from the center of the record (see drawing), what is the wavelength in the groove?

Short Answer

Expert verified
The wavelength is \( 6.98 \times 10^{-5} \, \text{m} \).

Step by step solution

01

Identify the Given Parameters

We're given the angular velocity \( \omega = 3.49 \, \text{rad/s} \) and the frequency of the tone \( f = 5.00 \, \text{kHz} = 5000 \, \text{Hz} \). The radius of the groove \( r = 0.100 \, \text{m} \). We need to find the wavelength \( \lambda \) in the groove.
02

Calculate Linear Velocity

The linear velocity \( v \) of the groove is found using the formula \( v = \omega \times r \). Substitute the given values: \[ v = 3.49 \, \text{rad/s} \times 0.100 \, \text{m} = 0.349 \, \text{m/s} \]
03

Compute the Wavelength

The wavelength \( \lambda \) can be computed using the relationship \( \lambda = \frac{v}{f} \). Substitute the values for \( v \) and \( f \):\[ \lambda = \frac{0.349 \, \text{m/s}}{5000 \, \text{Hz}} = 6.98 \times 10^{-5} \, \text{m} \]
04

Conclusion

Thus, the wavelength of the tone cut into the groove is \( 6.98 \times 10^{-5} \, \text{m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Velocity
Angular velocity describes how fast an object rotates or revolves around a central point. It is denoted by \( \omega \) and measured in radians per second (rad/s). Imagine a spinning record; angular velocity tells us how quickly it is spinning.

The relationship between angular velocity and linear velocity is crucial in many physics problems. For a point on a spinning disc, such as a record, the angular velocity remains constant if the speed of rotation doesn't change. Understanding this helps in converting angular velocity into other useful parameters, like linear velocity.

The formula used for angular velocity is \( \omega = \frac{\theta}{t} \), where \( \theta \) is the angle in radians and \( t \) is the time in seconds. However, in exercises, like the one provided, you might directly use the given angular speed to find linear velocity and other quantities.
Linear Velocity
Linear velocity is a straight-line speed of an object moving along a path. When considering rotation, it's the speed at which a point on the outer edge of a spinning object moves. Unlike angular velocity which deals with circular motion, linear velocity gives a measure that feels more intuitive, similar to speeds we encounter every day like kilometers per hour.

In rotational systems, such as a record's groove, linear velocity can be calculated from angular velocity and radius using the formula \( v = \omega \times r \). Here, \( \omega \) is the angular velocity and \( r \) is the radius or distance from the center of rotation to the point.

This derived velocity helps in bridging the gap between the concept of spinning and the linear distance something moves over time. Once you have the linear velocity, you can calculate other related quantities, like the wavelength of a sound in the groove of a record.
Frequency
Frequency refers to how often something happens over a certain period of time. In the context of sound waves or objects in rotation, frequency tells us the number of cycles or revolutions per second.

It is measured in hertz (Hz), which is equivalent to one cycle per second. For example, a frequency of 5,000 Hz implies there are 5,000 cycles occurring every second.

Understanding frequency is essential in determining wave properties such as wavelength. The relationship between linear velocity, frequency, and wavelength of a sound wave is crucial. The frequency of the sound in the record groove ties directly to the linear speed at which the sound wave travels through the medium.

Knowing the frequency and linear velocity, you can find the wavelength using the formula \( \lambda = \frac{v}{f} \), where \( \lambda \) is the wavelength, \( v \) is the linear velocity, and \( f \) is the frequency.
Physics Problem Solving
Physics problem solving often requires you to apply physical formulas and concepts to derive answers from given situations. It involves a few comprehensive steps to ensure accuracy.

  • Identify the Known Variables: Start by clearly identifying what you're given and what you need to find. This typically involves listing out all relevant parameters like velocity, frequency, or radius to keep track of necessary details.

  • Apply Appropriate Formulas: Next, apply the right formulas. In rotational dynamics, this involves using equations for angular velocity, linear velocity, and wave mechanics.

  • Calculate and Interpret: Perform calculations step-by-step, substituting in the known values to solve for the unknown. Make sure to include units in these calculations to avoid mistakes.

  • Analyze the Results: After computing, interpret the results to ensure they make sense physically. This is essential as a final check to confirm the correctness of your answer.


Mastering these problem-solving steps empowers you to tackle a variety of physics topics, turning seemingly complex problems into manageable steps.

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Most popular questions from this chapter

Two wires are parallel, and one is directly above the other. Each has a length of \(50.0 \mathrm{m}\) and a mass per unit length of \(0.020 \mathrm{kg} / \mathrm{m}\). However, the tension in wire \(\mathrm{A}\) is \(6.00 \times 10^{2} \mathrm{N},\) and the tension in wire \(\mathrm{B}\) is \(3.00 \times 10^{2} \mathrm{N} .\) Transverse wave pulses are generated simultaneously, one at the left end of wire \(A\) and one at the right end of wire \(B\). The pulses travel toward each other. How much time does it take until the pulses pass each other?

When Gloria wears her hearing aid, the sound intensity level increases by \(30.0 \mathrm{dB}\). By what factor does the sound intensity increase?

A car driving along a highway at a speed of \(23 \mathrm{m} / \mathrm{s}\) strays onto the shoulder. Evenly spaced parallel grooves called "rumble strips" are carved into the pavement of the shoulder. Rolling over the rumble strips causes the car's wheels to oscillate up and down at a frequency of 82 Hz. How far apart are the centers of adjacent rumble-strip grooves?

The mass of a string is \(5.0 \times 10^{-3} \mathrm{kg}\), and it is stretched so that the tension in it is \(180 \mathrm{N}\). A transverse wave traveling on this string has a frequency of \(260 \mathrm{Hz}\) and a wavelength of \(0.60 \mathrm{m} .\) What is the length of the string?

A recording engineer works in a soundproofed room that is \(44.0 \mathrm{dB}\) quieter than the outside. If the sound intensity that leaks into the room is \(1.20 \times 10^{-10} \mathrm{W} / \mathrm{m}^{2},\) what is the intensity outside?

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