the equation of transverse wave on a string is y = 92.0mm*sin(20.0 m^(-1))x - (600.0 s^(-1))t. the tension…

the equation of transverse wave on a string is y = 92.0mm*sin(20.0 m^(-1))x - (600.0 s^(-1))t. the tension in the string is f = 15.0 n. (a) what is the speed of the particle at x = 10.0 cm, t = 5.0 s? (b) what is the acceleration of a particle at x = 10.0 cm, t = 5.0 s? (c) find the linear density of the string? a) u = \\frac{\\partial y}{\\partial t}=\\frac{\\partial}{\\partial t}(a\\sin kx - \\omega t) u(x,t)= - a\\omega\\cos(kx - \\omega t) y(x,t)=a\\sin(kx\\pm\\omega t) y(x,t)=a\\sin kx + \\omega t - y(x,t)=92.0\\times10^{-3}m\\sin(\\frac{20.0}{1m}x - \\frac{600.0}{1s}t) u(x = 0.1m,t = 5.0s)= - 0.092\\times600\\frac{m}{s}\\cos\\frac{20.0\\times0.1m}{1m}-\\frac{600.0\\times5.0s}{1s}

the equation of transverse wave on a string is y = 92.0mm*sin(20.0 m^(-1))x - (600.0 s^(-1))t. the tension in the string is f = 15.0 n. (a) what is the speed of the particle at x = 10.0 cm, t = 5.0 s? (b) what is the acceleration of a particle at x = 10.0 cm, t = 5.0 s? (c) find the linear density of the string? a) u = \\frac{\\partial y}{\\partial t}=\\frac{\\partial}{\\partial t}(a\\sin kx - \\omega t) u(x,t)= - a\\omega\\cos(kx - \\omega t) y(x,t)=a\\sin(kx\\pm\\omega t) y(x,t)=a\\sin kx + \\omega t - y(x,t)=92.0\\times10^{-3}m\\sin(\\frac{20.0}{1m}x - \\frac{600.0}{1s}t) u(x = 0.1m,t = 5.0s)= - 0.092\\times600\\frac{m}{s}\\cos\\frac{20.0\\times0.1m}{1m}-\\frac{600.0\\times5.0s}{1s}

Answer

Explanation:

Step1: Identify the wave - function and its derivative for particle speed

The transverse wave function is $y(x,t)=A\sin(kx - \omega t)$ where $A = 92.0\times10^{-3}\ m$, $k = 20.0\ m^{-1}$, $\omega=600.0\ s^{-1}$. The speed of a particle on the string is given by $u=\frac{\partial y}{\partial t}$. Differentiating $y(x,t)$ with respect to $t$ using the chain - rule, we get $u(x,t)=-A\omega\cos(kx - \omega t)$.

Step2: Substitute the given values of $x$ and $t$

We are given $x = 10.0\ cm=0.1\ m$ and $t = 5.0\ s$. Substitute $A = 92.0\times10^{-3}\ m$, $\omega = 600.0\ s^{-1}$, $k = 20.0\ m^{-1}$, $x = 0.1\ m$ and $t = 5.0\ s$ into the formula for $u(x,t)$. [ \begin{align*} u(0.1,5)&=- (92.0\times10^{-3})\times600\cos((20\times0.1)-(600\times5))\ &=- 55.2\cos(2 - 3000)\ &=- 55.2\cos(-2998)\ &=- 55.2\cos(2998)\ &=- 55.2\cos(2998\bmod{2\pi})\ &=- 55.2\cos(1.92)\ \end{align*} ] where we use the fact that $\cos(x)=\cos(x + 2n\pi)$ for any integer $n$. $\cos(1.92)\approx - 0.27$. So $u(0.1,5)=-55.2\times(- 0.27)=14.9\ m/s$.

Step3: Find the acceleration of the particle

The acceleration of the particle is $a=\frac{\partial u}{\partial t}=\frac{\partial^{2}y}{\partial t^{2}}$. Differentiating $u(x,t)=-A\omega\cos(kx - \omega t)$ with respect to $t$ gives $a(x,t)=A\omega^{2}\sin(kx - \omega t)$. Substitute $A = 92.0\times10^{-3}\ m$, $\omega = 600.0\ s^{-1}$, $k = 20.0\ m^{-1}$, $x = 0.1\ m$ and $t = 5.0\ s$ into the formula for $a(x,t)$. [ \begin{align*} a(0.1,5)&=(92.0\times10^{-3})\times(600)^{2}\sin((20\times0.1)-(600\times5))\ &=92.0\times10^{-3}\times360000\sin(2 - 3000)\ &=33120\sin(-2998)\ &=33120\sin(2998)\ &=33120\sin(2998\bmod{2\pi})\ &=33120\sin(1.92)\ \end{align*} ] Since $\sin(1.92)\approx0.96$, $a(0.1,5)=33120\times0.96 = 31795.2\ m/s^{2}$.

Step4: Find the linear density of the string

The wave speed $v$ is related to the tension $F$ and linear density $\mu$ by $v=\sqrt{\frac{F}{\mu}}$, and the wave speed $v$ is also given by $v=\frac{\omega}{k}$. First, find $v=\frac{\omega}{k}=\frac{600}{20}=30\ m/s$. Then, from $v = \sqrt{\frac{F}{\mu}}$, we can solve for $\mu$. Squaring both sides gives $v^{2}=\frac{F}{\mu}$, so $\mu=\frac{F}{v^{2}}$. Substitute $F = 15.0\ N$ and $v = 30\ m/s$ into the formula, we get $\mu=\frac{15}{30^{2}}=\frac{15}{900}=0.0167\ kg/m$.

Answer:

(a) $14.9\ m/s$ (b) $31795.2\ m/s^{2}$ (c) $0.0167\ kg/m$