Question

l = 0.80 m and that is under tension of f = 325 n. wave amplitude is a = 0.40 mm and standing wave has n = 4 loops. (a) find the wavelength? (b) find the frequency? (c) what is maximum magnitude of the transverse velocity? a) antinode node node 2λ = l λ = l/2 = r c) u(x,t)= - aωcos(kx ± ωt + φ) |u(x,t)|max = aω = a·2πfn = a·2πf4 = 0.004×2π×? m/s b) λn = 2l/n λ4 = 2l/4 = l/2 fn = v/λn = √(f/μ)/λn f4 = 1/λ4·√(mg/ml) = 2/l(325×0.80/0.0025)^(1/2) f4 = hz

Explanation:

Step1: Find the wavelength

For a standing - wave with $n$ loops on a string of length $L$, the relationship between the wavelength $\lambda$ and the length of the string is $\lambda=\frac{2L}{n}$. Given $L = 0.80\ m$ and $n = 4$, we substitute these values into the formula.
$\lambda=\frac{2\times0.80}{4}$

Step2: Calculate the linear mass density $\mu$

The cross - sectional area of the string is $A = 0.40\ mm^{2}=0.40\times10^{- 6}\ m^{2}$. Assume the density of the material of the string is $
ho$ (not given in the problem, but if we consider the mass per unit length $\mu$ in terms of the tension $F$ and wave speed $v$, we can also use another approach). First, from $\lambda=\frac{2L}{n}$, we have $\lambda = 0.4\ m$.
The wave speed $v$ on a string under tension $F$ is given by $v=\sqrt{\frac{F}{\mu}}$. Also, $v = f\lambda$. But we can find the wave speed from the standing - wave relationship.

Step3: Find the frequency $f$

We know that $v=\sqrt{\frac{F}{\mu}}$. First, we need to find $\mu$. If we assume the mass of the string is $m$ and length is $L$, $\mu=\frac{m}{L}$. Since we are not given mass directly, we use the fact that for a standing wave $v = f\lambda$. From $\lambda=\frac{2L}{n}$, and $v=\sqrt{\frac{F}{\mu}}$. We can also use the formula $f=\frac{v}{\lambda}$.
However, if we just focus on the first part (finding wavelength), $\lambda=\frac{2L}{n}=\frac{2\times0.80}{4}=0.4\ m$.
For part (b), to find the frequency $f$, we know that $v=\sqrt{\frac{F}{\mu}}$. First, we find $\mu$. The cross - sectional area $A = 0.40\ mm^{2}=0.40\times10^{-6}\ m^{2}$. Let's assume the density of the material of the string is $
ho$. Then $\mu=
ho A$. But we can also use $v = f\lambda$. Since $v=\sqrt{\frac{F}{\mu}}$, and $\lambda$ is known from part (a).
For part (c), the transverse velocity of a wave is given by $u(x,t)=-A\omega\cos(kx\pm\omega t+\varphi)$. The maximum transverse velocity $u_{max}=A\omega$. Since $\omega = 2\pi f$, and we need to find $f$ first from $v = f\lambda$ and $v=\sqrt{\frac{F}{\mu}}$.

Answer:

(a) $\lambda = 0.4\ m$
(b) First, find the wave speed $v=\sqrt{\frac{F}{\mu}}$, where $\mu$ needs to be calculated from the cross - sectional area and density (not given fully in the problem statement). Then $f=\frac{v}{\lambda}$ with $\lambda = 0.4\ m$ from part (a).
(c) $u_{max}=A\omega=A\times2\pi f$. After finding $f$ from part (b), substitute $A = 0.40\ mm = 0.0004\ m$ and $\omega = 2\pi f$ to get the maximum transverse velocity.