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Please use the equations provided to solve The speed of sound in a liquid or solid it’s much faster than the speed of sound in air, which is around 340 m/s. Write the equation Y(x,t)for a sound wave with the frequency and an amplitude given below. Show wavelength, wave number, period and natural frequency. Numbers in decimals not pi. Steel with a speed of 6956 m/s Assume frequency of 2013 Hz and amplitude of 18.3 dB.

Please use the equations provided to solve The speed of sound in a liquid or solid it’s much faster than the speed of sound in air, which is around 340 m/s. Write the equation Y(x,t)for a sound wave with the frequency and an amplitude given below. Show wavelength, wave number, period and natural frequency. Numbers in decimals not pi. Steel with a speed of 6956 m/s Assume frequency of 2013 Hz and amplitude of 18.3 dB.

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Please use the equations provided to solve The speed of sound in a liquid or solid it’s much faster than the speed of sound in air, which is around 340 m/s. Write the equation Y(x,t)for a sound wave with the frequency and an amplitude given below. Show wavelength, wave number, period and natural frequency. Numbers in decimals not pi. Steel with a speed of 6956 m/s Assume frequency of 2013 Hz and amplitude of 18.3 dB.
**Vibrational Motion:** - \( T = \frac{1}{f} \) - \( \omega = 2\pi f \) **Wave Equation:** - \( v = f \lambda \) **Standing Waves:** - \( f_n = n f_0 \) - \( \lambda_n = \frac{2L}{n} \) **Wave Speed on a String:** - \( v = \sqrt{\frac{T}{\mu}} \) - \( \mu = \frac{\text{mass}}{\text{length}} \) **Travelling Wave:** - \( Y(x, t) = A \sin (kx - \omega t) \) **Explanation:** - **Vibrational Motion**: The period \( T \) is the reciprocal of frequency \( f \). The angular frequency \( \omega \) is given by \( 2\pi f \). - **Wave Equation**: The wave speed \( v \) is the product of frequency \( f \) and wavelength \( \lambda \). - **Standing Waves**: The frequency for a harmonic \( n \) is \( f_n = n f_0 \) and wavelength is given by \( \lambda_n = \frac{2L}{n} \), where \( L \) is the length of the medium. - **Wave Speed on a String**: The speed \( v \) depends on tension \( T \) and linear density \( \mu \) (mass per unit length). - **Travelling Wave**: The displacement \( Y(x, t) \) is described using amplitude \( A \) and wave numbers \( k \) and \( \omega \), representing spatial and temporal changes respectively.
Transcribed Image Text:**Vibrational Motion:** - \( T = \frac{1}{f} \) - \( \omega = 2\pi f \) **Wave Equation:** - \( v = f \lambda \) **Standing Waves:** - \( f_n = n f_0 \) - \( \lambda_n = \frac{2L}{n} \) **Wave Speed on a String:** - \( v = \sqrt{\frac{T}{\mu}} \) - \( \mu = \frac{\text{mass}}{\text{length}} \) **Travelling Wave:** - \( Y(x, t) = A \sin (kx - \omega t) \) **Explanation:** - **Vibrational Motion**: The period \( T \) is the reciprocal of frequency \( f \). The angular frequency \( \omega \) is given by \( 2\pi f \). - **Wave Equation**: The wave speed \( v \) is the product of frequency \( f \) and wavelength \( \lambda \). - **Standing Waves**: The frequency for a harmonic \( n \) is \( f_n = n f_0 \) and wavelength is given by \( \lambda_n = \frac{2L}{n} \), where \( L \) is the length of the medium. - **Wave Speed on a String**: The speed \( v \) depends on tension \( T \) and linear density \( \mu \) (mass per unit length). - **Travelling Wave**: The displacement \( Y(x, t) \) is described using amplitude \( A \) and wave numbers \( k \) and \( \omega \), representing spatial and temporal changes respectively.
Expert Solution
Step 1

Given: The speed of sound is 340 m/s. 

           The frequency is 2013 Hz. 

           The amplitude is 18.3 dB. 

           The steel with sound is 6956 m/s. 

To determine: The equation Y(x,t) 

The speed of sound is 

v=fλ

where v is the speed, f is the frequency and λ is the wavelength 

Substitute 6956 m/s for v and 2013 Hz for f 

6956 m/s=2013 Hz×λλ=3.45 m

The wavelength is 3.45 m. 

The time period is 

T=1fT=12013 HzT=4.97×10-4 s

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