Question

A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation $$y(x,t)=(8.50\mathrm{mm})\cos (172\mathrm{rad}/\mathrm{m}x-4830\mathrm{rad}/\mathrm{s}t)$$ Assume that the tension of the string is constant and equal to $W$. (a) How much time does it take a pulse to travel the full length of the string? (b) What is the weight $W$? (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for waves traveling $down$ the string?

Short answer

(a) 0.0534 s, (b) 6.565 N, (c) 41 wavelengths, (d) $y(x,t)=(8.50\text{mm})\cos (172\text{rad/m}x+4830\text{rad/s}t)$.

Step-by-step solution

Identify Given Information

We know the length of the string, $L=1.50$ m, and the weight of the string is 0.0125 N. The wave equation for the string is given by $y(x,t)=(8.50\text{mm})\cos (172\text{rad/m}x-4830\text{rad/s}t)$. From the wave equation, we identify the wave number $k=172\text{rad/m}$ and the angular frequency $\omega =4830\text{rad/s}$.

Calculate Wave Speed (a)

The wave speed $v$ can be derived from the wave equation using $v=\frac{\omega }{k}$. Given $\omega =4830\text{rad/s}$ and $k=172\text{rad/m}$, we calculate:$$v=\frac{4830}{172}=28.081\text{m/s}$$

Time for a Pulse to Travel Full Length (a)

The time $T$ it takes for a pulse to travel the full length of the string is given by $T=\frac{L}{v}$. Using $L=1.50\text{m}$ and $v=28.081\text{m/s}$:$$T=\frac{1.50}{28.081}\approx 0.0534\text{s}$$

Calculate Tension and Weight (b)

The tension in the string is equal to the weight $W$. The wave speed can also be given by $v=\sqrt{\frac{T}{\mu }}$, where $\mu$ is the linear mass density of the string. To find $\mu$, use the weight of the string and its length: $\mu =\frac{0.0125}{1.50}=0.00833\text{kg/m}$.Solve for $T$:$$v^2=\frac{T}{\mu }$$$$T=v^2\mu =(28.081{)}^2\times 0.00833\approx 6.565\text{N}$$Hence, $W=6.565\text{N}$.

Wavelength on the String (c)

The wavelength $\lambda$ is given by $\lambda =\frac{2\pi }{k}$. Plug in $k=172\text{rad/m}$:$$\lambda =\frac{2\pi }{172}\approx 0.0365\text{m}$$To find how many wavelengths are on the string, use $L=1.50\text{m}$:$$n=\frac{L}{\lambda }=\frac{1.50}{0.0365}\approx 41.1$$Therefore, approximately 41 full wavelengths fit on the string.

Equation for Waves Traveling Down the String (d)

For waves traveling down the string, the wave number remains the same, but the sign of the angular frequency changes:$$y(x,t)=(8.50\text{mm})\cos (172\text{rad/m}x+4830\text{rad/s}t)$$

Key concepts

Wave Speed

Wave speed is a fundamental concept in wave mechanics that describes how quickly a wave can travel through a medium. In simple terms, it tells us how fast the energy or information carried by the wave moves from one point to another.

To calculate the wave speed ($v$), we can use information from the wave equation. The formula to find wave speed is:

• $$v=\frac{\omega }{k}$$

where $\omega$ is the angular frequency, and $k$ is the wave number.

In the original exercise, the angular frequency is 4830 rad/s and the wave number is 172 rad/m. By applying our formula, the wave speed becomes:

• $$v=\frac{4830}{172}=28.081\text{m/s}$$

This calculation shows us how fast the wave pulse moves along the string.

Wavelength

A wavelength is an essential concept in understanding waves. It's the distance over which the wave's shape repeats. You can visualize it as the length from crest to crest or trough to trough of the wave.

We can find the wavelength ($\lambda$) using the wave number $k$. The relationship between them is given by the formula:

• $$\lambda =\frac{2\pi }{k}$$

Given $k=172$ rad/m:

• $$\lambda =\frac{2\pi }{172}\approx 0.0365\text{m}$$

This means that each wavelength is approximately 0.0365 meters long.

The original exercise also finds how many wavelengths fit on the string. For a string of length 1.50 meters:

• $$n=\frac{1.50}{0.0365}\approx 41.1$$

So, about 41 waves fit perfectly along the length of the string at any moment.

Tension in String

The tension in a string is a force that stretches it, playing a crucial role in determining how waves travel along it. In physical terms, the tension keeps the string tight and affects the speed of wave propagation in the string.

To connect tension ($T$), wave speed ($v$), and linear mass density ($\mu$) of the string, we use the equation:

• $$v=\sqrt{\frac{T}{\mu }}$$

The linear mass density, $\mu$, is the mass per unit length and in this problem is 0.00833 kg/m.

Rearranging to find tension $T$ gives:

• $$T=v^2\mu$$

Plug in the values from the original exercise ($v=28.081\text{m/s},\mu =0.00833\text{kg/m}$):

• $$T=(28.081{)}^2\times 0.00833\approx 6.565\text{N}$$

Hence, the tension is approximately 6.565 N, equivalent to the supporting weight $W$, ensuring the string operates under constant tension.

Wave Equation

The wave equation is a mathematical representation that describes the characteristics of wave behavior along a medium.

For our string scenario, the provided wave equation is:

• $y(x,t)=(8.50\text{mm})\cos (172\text{rad/m}x-4830\text{rad/s}t)$

This equation tells us the wave's movement along the string over time.

Different elements in the wave equation include:

• The amplitude (8.50 mm), representing the maximum displacement of wave particles from their rest position.
• The wave number (172 rad/m), which is linked to wavelength as we discussed.
• The angular frequency (4830 rad/s), showing how rapidly the wave oscillates.

For waves traveling downwards, the formula is adjusted slightly:

• $y(x,t)=(8.50\text{mm})\cos (172\text{rad/m}x+4830\text{rad/s}t)$

Here, the sign of the angular frequency changes, indicating a different direction for the wave.