a string has linear density μ = 525 g/m and is under tension f = 45 n. we send sinusoidal wave with…

a string has linear density μ = 525 g/m and is under tension f = 45 n. we send sinusoidal wave with frequency f = 120 hz and amplitude a = 8.5 cm along the string. (a)find the speed of the wave. (b)find the displacement function of a particle of string. (c)at what average rate does the wave transport energy. (d)find the intensity of this wave at a point r = 15 m from the source. a) v = λf : mechanical v = √(f/μ) f = 45 n μ = 0.525 kg/m } v = √(45/0.525) =? m/s b) y(x,t) = a sin (kx - ωt) a = 0.085 m k = 2π/λ = 2πf/v = 2π×120/v m⁻¹ =? m⁻¹ ω = 2πf = 2π×120 rad/s =? y(x,t) = 0.085m sin (? m⁻¹ x -? s⁻¹ t) · linear mass density · wave speed of a stretched string · energy and power in a wave · intensity pavg = e/t = 1/2μa²ω²λ/t intensity, i = p/a

a string has linear density μ = 525 g/m and is under tension f = 45 n. we send sinusoidal wave with frequency f = 120 hz and amplitude a = 8.5 cm along the string. (a)find the speed of the wave. (b)find the displacement function of a particle of string. (c)at what average rate does the wave transport energy. (d)find the intensity of this wave at a point r = 15 m from the source. a) v = λf : mechanical v = √(f/μ) f = 45 n μ = 0.525 kg/m } v = √(45/0.525) =? m/s b) y(x,t) = a sin (kx - ωt) a = 0.085 m k = 2π/λ = 2πf/v = 2π×120/v m⁻¹ =? m⁻¹ ω = 2πf = 2π×120 rad/s =? y(x,t) = 0.085m sin (? m⁻¹ x -? s⁻¹ t) · linear mass density · wave speed of a stretched string · energy and power in a wave · intensity pavg = e/t = 1/2μa²ω²λ/t intensity, i = p/a

Answer

Explanation:

Step1: Calculate wave speed

The wave - speed formula for a stretched string is $v=\sqrt{\frac{F}{\mu}}$. Given $F = 45\ N$ and $\mu=525\ g/m=0.525\ kg/m$. $v=\sqrt{\frac{45}{0.525}}\approx\sqrt{85.71}\approx9.26\ m/s$

Step2: Calculate wave number $k$

The wave number $k=\frac{2\pi}{\lambda}$, and since $v = f\lambda$, then $\lambda=\frac{v}{f}$. So $k=\frac{2\pi f}{v}$. Given $f = 120\ Hz$ and $v\approx9.26\ m/s$. $k=\frac{2\pi\times120}{9.26}\approx\frac{753.98}{9.26}\approx81.42\ m^{-1}$ The angular frequency $\omega = 2\pi f=2\pi\times120 = 240\pi\ rad/s$ The displacement - function of a sinusoidal wave is $y(x,t)=A\sin(kx-\omega t)$. Given $A = 8.5\ cm = 0.085\ m$. So $y(x,t)=0.085\sin(81.42x - 240\pi t)$

Step3: Calculate the average power of the wave

The average power of a wave is $P_{avg}=\frac{1}{2}\mu A^{2}\omega^{2}v$. Substitute $\mu = 0.525\ kg/m$, $A = 0.085\ m$, $\omega=240\pi\ rad/s$, and $v\approx9.26\ m/s$ $\omega^{2}=(240\pi)^{2}\approx(240\times3.14)^{2}\approx(753.6)^{2}\approx567912.96$ $P_{avg}=\frac{1}{2}\times0.525\times(0.085)^{2}\times567912.96\times9.26$ $P_{avg}=\frac{1}{2}\times0.525\times0.007225\times567912.96\times9.26$ $P_{avg}\approx\frac{1}{2}\times0.525\times0.007225\times5268974.01$ $P_{avg}\approx\frac{1}{2}\times0.525\times38081.94$ $P_{avg}\approx9995.97\ W$

Step4: Calculate the intensity of the wave

The intensity of a wave is $I=\frac{P_{avg}}{4\pi r^{2}}$. Given $r = 15\ m$ and $P_{avg}\approx9995.97\ W$ $I=\frac{9995.97}{4\pi\times15^{2}}=\frac{9995.97}{4\pi\times225}=\frac{9995.97}{2827.43}\approx3.53\ W/m^{2}$

Answer:

(a) $v\approx9.26\ m/s$ (b) $y(x,t)=0.085\sin(81.42x - 240\pi t)$ (c) $P_{avg}\approx9995.97\ W$ (d) $I\approx3.53\ W/m^{2}$