Transverse waves on a string have a velocity of 15 m/s, amplitude of 0.10m and a wavelength of 25 cm. The waves travel in the +x direction and at t=0 the x=0 end of the string is at a maximum upward displacement. Find the transverse displacement at x=.2m and t=0.3s (m):
Properties of sound
A sound wave is a mechanical wave (or mechanical vibration) that transit through media such as gas (air), liquid (water), and solid (wood).
Quality Of Sound
A sound or a sound wave is defined as the energy produced due to the vibrations of particles in a medium. When any medium produces a disturbance or vibrations, it causes a movement in the air particles which produces sound waves. Molecules in the air vibrate about a certain average position and create compressions and rarefactions. This is called pitch which is defined as the frequency of sound. The frequency is defined as the number of oscillations in pressure per second.
Categories of Sound Wave
People perceive sound in different ways, like a medico student takes sound as vibration produced by objects reaching the human eardrum. A physicist perceives sound as vibration produced by an object, which produces disturbances in nearby air molecules that travel further. Both of them describe it as vibration generated by an object, the difference is one talks about how it is received and other deals with how it travels and propagates across various mediums.
![### Problem Statement: Transverse Wave Motion on a String
Transverse waves on a string have a velocity of 15 m/s, an amplitude of 0.10 m, and a wavelength of 25 cm. The waves travel in the +x direction, and at \( t = 0 \), the \( x = 0 \) end of the string is at a maximum upward displacement.
**Task:**
Find the transverse displacement at \( x = 0.2 \) m and \( t = 0.3 \) s (in meters).
**Given Parameters:**
- Velocity (\( v \)) = 15 m/s
- Amplitude (\( A \)) = 0.10 m
- Wavelength (\( \lambda \)) = 25 cm (0.25 m)
- Initial conditions: At \( t = 0 \), \( x = 0 \) is at maximum upward displacement.
**Solution Approach:**
The equation of a transverse wave traveling in the +x direction can be written as:
\[ y(x,t) = A \cos\left( \frac{2\pi}{\lambda}(x - vt) \right) \]
**Substitute the values:**
\[ A = 0.10 \, \text{m} \]
\[ \lambda = 0.25 \, \text{m} \]
\[ v = 15 \, \text{m/s} \]
\[ x = 0.2 \, \text{m} \]
\[ t = 0.3 \, \text{s} \]
Thus,
\[ y(0.2, 0.3) = 0.10 \cos\left( \frac{2\pi}{0.25}(0.2 - 15 \cdot 0.3) \right) \]
Calculate the argument:
\[ 0.2 - 4.5 = -4.3 \]
Then,
\[ \frac{2\pi}{0.25} \times (-4.3) = 8\pi \times (-4.3) = -34.4\pi \]
\[ y(0.2, 0.3) = 0.10 \cos(-34.4\pi) \]
Since cosine is an even function:
\[ \cos(-34.4\pi) = \cos(34.4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdfe80350-6fa1-469d-aae2-95b4dd3aa499%2F93f63088-e027-42e2-bc00-4c30f3a3f41c%2Fsdutdb_processed.png&w=3840&q=75)
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