The equation of a transverse wave traveling along a very long string is y=9.24 sin(0.0317x+ 3.48t), where x and y are expressed in centimeters and t is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave and (f) the maximum transverse speed of a particle in the string. (g) What is the transverse displacement at x 5.98 cm when t = 0.792 s?

College Physics
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Chapter1: Units, Trigonometry. And Vectors
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### Transverse Wave Analysis

The equation of a transverse wave traveling along a very long string is given by:
\[ y = 9.24 \sin(0.0317\pi x + 3.48\pi t) \]

where \( x \) and \( y \) are expressed in centimeters (cm) and \( t \) is in seconds (s). We need to determine the following parameters:

#### (a) Amplitude
**Given**: 9.24 cm

The amplitude of the wave is the maximum displacement from the equilibrium position, which here is 9.24 cm.

#### (b) Wavelength
**Calculated Value**: 63.0915 cm

The wavelength (\(\lambda\)) can be determined from the wave number \(k\).

\[ k = 0.0317\pi \]
\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{0.0317\pi} \approx 63.0915 \text{ cm} \]

#### (c) Frequency
**Calculated Value**: 1.74 rad/s or s\(^{-1}\)

The angular frequency (\(\omega\)) is given by:
\[ \omega = 3.48\pi \]

The frequency (\(f\)) is:
\[ f = \frac{\omega}{2\pi} = \frac{3.48\pi}{2\pi} = 1.74 \text{ rad/s or s}^{-1} \]

#### (d) Speed
**Calculated Value**: 109.779 cm/s

The speed (\(v\)) of the wave can be calculated using:
\[ v = \lambda \cdot f \]

Given:
\[ \lambda = 63.0915 \text{ cm} \]
\[ f = 1.74 \text{ s}^{-1} \]
\[ v = 63.0915 \text{ cm} \times 1.74 \text{ s}^{-1} \approx 109.779 \text{ cm/s} \]

#### (e) Direction of Propagation
**Given**: -x direction

The direction of propagation is determined by the sign in front of the wave number term in the equation. Here, it moves in the -x direction.

#### (f) Maximum Transverse Speed of a Particle in the String
Transcribed Image Text:### Transverse Wave Analysis The equation of a transverse wave traveling along a very long string is given by: \[ y = 9.24 \sin(0.0317\pi x + 3.48\pi t) \] where \( x \) and \( y \) are expressed in centimeters (cm) and \( t \) is in seconds (s). We need to determine the following parameters: #### (a) Amplitude **Given**: 9.24 cm The amplitude of the wave is the maximum displacement from the equilibrium position, which here is 9.24 cm. #### (b) Wavelength **Calculated Value**: 63.0915 cm The wavelength (\(\lambda\)) can be determined from the wave number \(k\). \[ k = 0.0317\pi \] \[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{0.0317\pi} \approx 63.0915 \text{ cm} \] #### (c) Frequency **Calculated Value**: 1.74 rad/s or s\(^{-1}\) The angular frequency (\(\omega\)) is given by: \[ \omega = 3.48\pi \] The frequency (\(f\)) is: \[ f = \frac{\omega}{2\pi} = \frac{3.48\pi}{2\pi} = 1.74 \text{ rad/s or s}^{-1} \] #### (d) Speed **Calculated Value**: 109.779 cm/s The speed (\(v\)) of the wave can be calculated using: \[ v = \lambda \cdot f \] Given: \[ \lambda = 63.0915 \text{ cm} \] \[ f = 1.74 \text{ s}^{-1} \] \[ v = 63.0915 \text{ cm} \times 1.74 \text{ s}^{-1} \approx 109.779 \text{ cm/s} \] #### (e) Direction of Propagation **Given**: -x direction The direction of propagation is determined by the sign in front of the wave number term in the equation. Here, it moves in the -x direction. #### (f) Maximum Transverse Speed of a Particle in the String
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