A wave on a string is described by D(x,t)=(2.00 cm)×sin⁡[(12.57 rad/m)x−(638 rad/s)t]D(x,t) = (2.00 \, \text{cm}) \times \sin[(12.57 \, \text{rad/m})x - (638 \, \text{rad/s}) t], where xx is in mm and tt in ss. The linear density of the string is 5.00 g/m5.00\text{ g/m}. What are The maximum speed of a point on the string?

Ch 16: Traveling Waves

Knight Calc - Physics for Scientists and Engineers 5th Edition

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Knight Calc 5th Edition

Ch 16: Traveling Waves

Problem 56c

Chapter 16, Problem 56c

A wave on a string is described by $D (x, t) = (2.00 cm) \times \sin [(12.57 rad/m) x - (638 rad/s) t]$ , where $x$ is in $m$ and $t$ in $s$ . The linear density of the string is $5.00 g slash m$ . What are The maximum speed of a point on the string?

Verified step by step guidance

Step 1: Identify the general equation for the wave on the string, which is given as D(x, t) = (2.00 cm) ✕ sin[(12.57 rad/m)x ─ (638 rad/s)t]. Here, the amplitude of the wave is A = 2.00 cm, the angular frequency is ω = 638 rad/s, and the wave number is k = 12.57 rad/m.

Step 2: Recall that the maximum speed of a point on the string occurs when the displacement D(x, t) is changing most rapidly. This is given by the derivative of D(x, t) with respect to time, which is the velocity of the point on the string: v(x, t) = ∂D(x, t)/∂t.

Step 3: Differentiate D(x, t) with respect to time t. Using the chain rule, ∂D(x, t)/∂t = Aω ✕ cos(kx ─ ωt). The maximum speed occurs when cos(kx ─ ωt) = ±1, which gives the maximum value of the velocity as v_max = Aω.

Step 4: Substitute the values of A and ω into the formula for v_max. Convert the amplitude A from cm to meters: A = 2.00 cm = 0.0200 m. The angular frequency ω is already given as 638 rad/s. Thus, v_max = Aω = (0.0200 m)(638 rad/s).

Step 5: The maximum speed of a point on the string is determined by the product of the amplitude and angular frequency. Perform the multiplication to find the numerical value if needed, but the formula v_max = Aω is the key result.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Wave Equation

The wave equation describes the behavior of waves, including their amplitude, frequency, and wavelength. In the given equation D(x,t) = (2.00 cm) × sin[(12.57 rad/m)x - (638 rad/s)t], the amplitude is 2.00 cm, the wave number is 12.57 rad/m, and the angular frequency is 638 rad/s. Understanding these parameters is essential for analyzing wave properties and their effects on the medium.

Maximum Speed of a Point on the String

The maximum speed of a point on a wave is determined by the product of the angular frequency and the amplitude of the wave. It can be calculated using the formula v_max = ωA, where ω is the angular frequency and A is the amplitude. This concept is crucial for determining how quickly a point on the string moves as the wave propagates.

Linear Density

Linear density is defined as the mass per unit length of a string, typically expressed in grams per meter (g/m). In this case, the linear density of the string is 5.00 g/m. This property affects the wave speed and tension in the string, influencing how the wave propagates through the medium and the energy carried by the wave.

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Related Practice

Textbook Question

FIGURE P16.57 shows a snapshot graph of a wave traveling to the right along a string at 45 m/s. At this instant, what is the velocity of points 1, 2, and 3 on the string?

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Textbook Question

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Textbook Question

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Textbook Question

A string that is under 50.0 N of tension has linear density 5.0 g/m. A sinusoidal wave with amplitude 3.0 cm and wavelength 2.0 m travels along the string. What is the maximum speed of a particle on the string?

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Textbook Question

The string in FIGURE P16.59 has linear density μ. Find an expression in terms of M, μ, and θ for the speed of waves on the string.

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Textbook Question

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