Problem 66 - University Physics with Modern Physics

Question

A strong string of mass 3.00 g and length 2.20 m is tied to supports at each end and is vibrating in its fundamental mode. The maximum transverse speed of a point at the middle of the string is 9.00 m/s. The tension in the string is 330 N. (a) What is the amplitude of the standing wave at its antinode? (b) What is the magnitude of the maximum transverse acceleration of a point at the antinode?

Step-by-Step Solution

Answer

(a) The amplitude is approximately 0.0127 m. (b) The maximum acceleration is about 805.7 m/s².

Step 1: Understanding the Fundamental Frequency

In the fundamental mode of vibration, the string resonates with one antinode at the center and nodes at each end. The length of the string is half of the wavelength of the standing wave. Thus, the wavelength \( \lambda \) is equal to twice the length of the string: \[ \lambda = 2 \times 2.20 \, \text{m} = 4.40 \, \text{m} \]

Step 2: Calculate the Wave Speed

The wave speed \( v \) can be calculated using the formula: \[ v = \sqrt{\frac{T}{\mu}} \] where \( T \) is the tension in the string \( 330 \, \text{N} \) and \( \mu \) is the linear mass density. The linear mass density \( \mu \) is the mass of the string per unit length: \[ \mu = \frac{3.00 \, \text{g}}{2.20 \, \text{m}} = \frac{0.003 \, \text{kg}}{2.20 \, \text{m}} \] Calculating \( \mu \): \[ \mu \approx 1.364 \times 10^{-3} \, \text{kg/m} \] Then the wave speed \( v \) is: \[ v = \sqrt{\frac{330}{1.364 \times 10^{-3}}} \approx 493.9 \, \text{m/s} \]

Step 3: Calculate the Frequency

Using the relationship between wave speed, wavelength, and frequency: \[ v = f \lambda \] Substitute the values of \( v \) and \( \lambda \): \[ 493.9 = f \times 4.40 \] Solve for \( f \): \[ f \approx 112.3 \, \text{Hz} \]

Step 4: Calculate the Amplitude at the Antinode

The maximum transverse velocity \( v_{\text{max}} \) at the antinode is related to the amplitude \( A \) and the angular frequency \( \omega \): \[ v_{\text{max}} = A \omega \] The angular frequency \( \omega \) is \( 2 \pi f \): \[ \omega = 2 \pi \times 112.3 \] So, the amplitude \( A \) is: \[ A = \frac{v_{\text{max}}}{\omega} = \frac{9}{2 \pi \times 112.3} \approx 0.0127 \, \text{m} \]

Step 5: Maximum Transverse Acceleration at the Antinode

The maximum transverse acceleration \( a_{\text{max}} \) is given by: \[ a_{\text{max}} = A \omega^2 \] Substitute the amplitude \( A \) from Step 4 and \( \omega \): \[ a_{\text{max}} = 0.0127 \times (2 \pi \times 112.3)^2 \] Calculating the value gives: \[ a_{\text{max}} \approx 805.7 \, \text{m/s}^2 \]

Key Concepts

Fundamental Frequency

The fundamental frequency is the lowest frequency at which a system naturally vibrates. For a string fixed at both ends, like in this exercise, the string vibrates with a pattern that has one antinode in the center and nodes at either end. This is considered the fundamental mode of vibration. Here, the string's length represents half of the wavelength, which is crucial in calculating other parameters like frequency and speed of the wave.

In this context, the wavelength \( \lambda \) is twice the physical length of the string. Therefore, it plays a significant role in determining the frequency \( f \) of vibration, which is calculated using \( f = \frac{v}{\lambda} \), where \( v \) is the wave speed. Understanding this fundamental mode is essential for analyzing the wave behaviors and patterns on strings in physics.

Wave Speed

Wave speed is a vital concept when studying waves on a string. It tells us how fast a wave propagates along the medium. For a wave on a stretched string, wave speed \( v \) is derived from the tension force \( T \) applied to the string and the string's linear mass density \( \mu \). The formula is \( v = \sqrt{\frac{T}{\mu}} \).

This formula shows that the wave speed increases with greater tension in the string or lesser mass per length unit. In this exercise, the tension is given as 330 N, and you've calculated the linear mass density as \( 1.364 \times 10^{-3} \text{ kg/m} \). Plugging these values into the equation gives a significant wave speed of approximately 493.9 m/s.

By understanding wave speed in conjunction with fundamental frequency, you gain insight into how waves behave and the influence of different physical properties on the wave's movement.

Angular Frequency

Angular frequency is an extension of the concept of frequency into circular or harmonic motion. It tells us how rapidly the wave oscillates in radians per second and is a key component when analyzing waves and vibrations. It is calculated as \( \omega = 2\pi f \), where \( f \) is the wave frequency.

In the exercise, the frequency \( f \) was calculated to be approximately 112.3 Hz. To find the angular frequency, you multiply this by \( 2\pi \), resulting in \( \omega \approx 705.6 \text{ rad/s} \).

The angular frequency connects to various other wave properties such as amplitude and maximum acceleration. It shows the rotational equivalent of regular frequency, offering deep insights into the wave's dynamical behavior.

Linear Mass Density

Linear mass density \( \mu \) represents how much mass is distributed along a unit length of a string or other one-dimensional medium. It is expressed in \( \text{kg/m} \) and is a crucial factor in determining the wave speed and behavior on the string.

For a string, \( \mu \) can be easily calculated by dividing the mass of the string by its length. In this exercise, the string's mass is given as 3.00 g (or 0.003 kg) and its length as 2.20 m, yielding \( \mu = \frac{0.003}{2.20} \approx 1.364 \times 10^{-3} \text{ kg/m} \).

Linear mass density affects wave speed as seen in the wave speed formula. The smaller the \( \mu \), the faster waves travel along the string when under the same tension. This density concept is pivotal when designing or studying structures involving wave propagation such as musical instruments or scientific experiments.