Problem 96 - Physics

Question

A \(3.49-\mathrm{rad} / \mathrm{s}\left(33 \frac{1}{3} \mathrm{rpm}\right)\) 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?

Step-by-Step Solution

Answer

The wavelength in the groove is approximately \(6.98 \times 10^{-5} \mathrm{~m}.\)

Step 1: Convert Angular Velocity

Convert the angular velocity from radians per second to revolutions per second. Given that the angular velocity of the record is \(3.49 \mathrm{~rad/s}\), and knowing that one revolution is \(2\pi\) radians, calculate the revolutions per second (rps) as follows:\[\text{Revolutions per second} = \frac{3.49}{2\pi}\]

Step 2: Calculate Record's Linear Speed

To find the linear speed of the groove, use the formula for linear speed, \(v = r \cdot \omega\), where \(r\) is the radius (0.100 m from the center) and \(\omega\) is the angular velocity. Calculate the speed:\[ v = 0.100 \times 3.49 = 0.349 \mathrm{~m/s}\]

Step 3: Calculate Wavelength in Groove

To find the wavelength of the sound in the record groove, divide the linear speed of the record by the frequency of the tone. The formula is \(\lambda = \frac{v}{f}\), where \(v\) is the linear speed (0.349 m/s) and \(f\) is the frequency of the tone (5000 Hz):\[\lambda = \frac{0.349}{5000} = 6.98 \times 10^{-5} \mathrm{~m}\]

Key Concepts

Angular Velocity

Angular velocity is a key concept in understanding rotational motion. It tells us how fast an object rotates or spins around an axis.
For example, in our exercise, the record has an angular velocity of 3.49 radians per second. This means that every second, the record turns through an angle of 3.49 radians.

This unit, radians per second (\(\mathrm{rad/s}\)), is often converted into revolutions per minute (rpm) or revolutions per second (rps) when needed.

• 1 revolution equals \(2\pi\) radians.
• To convert from \(\mathrm{rad/s}\) to rps, divide the angular velocity by \(2\pi\).

This concept is crucial when dealing with objects like wheels, fans, and records that involve circular motion.

Linear Speed

Linear speed refers to the speed at which a point travels along its path. In the case of circular motion, it is the speed along the circumference of the circle.
In our example, we calculated linear speed using the formula \(v = r \cdot \omega\), where \(r\) is the distance from the center, or radius, and \(\omega\) is the angular velocity.

This formula transforms rotational motion into a straight distance per time. For our record, the linear speed was calculated as:

• \(v = 0.100 \ \mathrm{m} \times 3.49 \ \mathrm{rad/s} = 0.349 \ \mathrm{m/s}\)

Understanding linear speed is important in many aspects, including determining the real-world speed of any rotating object.

Wavelength Calculation

Wavelength is a measurement that describes the distance between two identical points on a wave, like from crest to crest.
In the exercise, the problem was to find the wavelength of a 5 kHz tone as it moves around the groove of a record at a certain speed.

We use the formula \(\lambda = \frac{v}{f}\), where:

• \(\lambda\) is the wavelength,
• \(v\) is the linear speed (calculated as 0.349 m/s),
• \(f\) is the frequency (5000 Hz).

Applying these values:

• \(\lambda = \frac{0.349}{5000} = 6.98 \times 10^{-5} \ \mathrm{m}\)

It's essential to understand wavelength as it helps us explore how different frequencies behave in various mediums.

Frequency

Frequency is the number of times an event repeats per unit time. In wave physics, it measures how often the wave crest passes a certain point.
The unit of frequency is Hertz (Hz), which represents cycles per second.

In the given problem, the frequency of the tone is 5 kHz or 5000 Hz. This implies that the sound wave's complete cycles occur 5000 times each second.

• Higher frequency means more wave cycles in a second.
• Frequency is inversely related to wavelength; as frequency increases, the wavelength decreases.

Understanding frequency is crucial for comprehending various phenomena, like sound tone, color of light, and even signal transmissions.

Circular Motion

Circular motion occurs when an object moves in a circular path. This involves understanding both angular and linear properties of motion.
One essential aspect is how linear speed and angular velocity relate to the path's radius.

In our exercise, the circular motion of the record allowed us to calculate both the linear speed and the eventual wavelength of the sound wave.

• Objects in circular motion experience constant change in direction.
• This requires a centripetal force directed toward the center of the circle.
• The linear speed increases with the radius when angular velocity remains constant.

Learning about circular motion helps us understand real-life applications such as satellites orbiting planets, cars rounding a curve, and, of course, records spinning on turntables.