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Chapter 16: Problem 10

10\. Speaker Diaphragm A loudspeaker diaphragm is oscillating in simple harmonic motion with a frequency of \(440 \mathrm{~Hz}\) and \(\mathrm{a}\) maximum displacement of \(0.75 \mathrm{~mm}\). What are (a) the angular frequency, (b) the maximum speed and (c) the magnitude of the maximum acceleration?

Short Answer

Expert verified
The angular frequency is 2764 rad/s, maximum speed is 2.073 m/s, and maximum acceleration is 5728.8 m/s^2.

Step by step solution

01

Calculate Angular Frequency

The angular frequency \( \omega \) is given by the formula \( \omega = 2 \pi f \), where \( f \) is the frequency. Given \( f = 440 \mathrm{Hz} \),\ \(\omega = 2 \pi \times 440 \mathrm{Hz} \ = 2764 \mathrm{rad/s} \).
02

Calculate Maximum Speed

The maximum speed \(v_{max}\ \) can be calculated using the formula \(v_{max} = \omega A \), where \(A\ \) is the maximum displacement. Given \(A \ = 0.75 \mathrm{mm} = 0.00075 \mathrm{m} \) and \(\omega = 2764 \mathrm{rad/s} \),\ \(\v_{max} = 2764 \mathrm{rad/s} \times 0.00075 \mathrm{m} \ = 2.073 \mathrm{m/s} \).
03

Calculate Magnitude of Maximum Acceleration

The magnitude of the maximum acceleration \(a_{max}\ \) can be calculated using the formula \(a_{max} = \omega^2 A \). Given \(A = 0.00075 \mathrm{m} \) and \(\omega = 2764 \mathrm{rad/s} \),\, \(\a_{max} = 2764^2 \mathrm{rad}^2/ \mathrm{s^2} \times 0.00075 \mathrm{m} \ = 5728.8 \mathrm{m/s^2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

angular frequency
Angular frequency is an important concept in understanding simple harmonic motion, like the oscillation of a loudspeaker diaphragm. Simple harmonic motion typically involves a cycle repeating over time. In this context, angular frequency (胃蠅) is the rate at which this cycle repeats in radians per second.

To find the angular frequency, you use the formula: 胃蠅 = 2蟺f.

Here, 'f' is the frequency of the oscillation. For example, if you have a frequency (f) of 440 Hz, you substitute it into the formula to get 胃蠅 = 2蟺 脳 440 Hz = 2764 rad/s.

This tells you that the diaphragm completes 2764 radians of oscillation each second!
maximum speed
The concept of maximum speed in harmonic motion helps us understand how fast the oscillating object moves at its fastest point. The maximum speed, denoted as (vm_max), is achieved when the object passes through its equilibrium position.

You can calculate maximum speed using the formula vm_max = 胃蠅A.

Here, 'A' is the maximum displacement, and 胃蠅 is the angular frequency. In our example, with an angular frequency (胃蠅) of 2764 rad/s and a maximum displacement (A) of 0.75 mm (which is 0.00075 meters), the maximum speed is calculated as:

vmax = 2764 rad/s 脳 0.00075 m = 2.073 m/s.

This means the diaphragm moves at a maximum speed of 2.073 meters per second!
maximum acceleration
Maximum acceleration in harmonic motion is another key concept, which helps us understand how quickly the velocity of the oscillating object changes. The magnitude of maximum acceleration (anamax) is experienced at the points of maximum displacement.

The formula to find maximum acceleration is: anamax = 胃蠅虏A.

For a diaphragm with an angular frequency (胃蠅) of 2764 rad/s and a maximum displacement (A) of 0.75 mm (or 0.00075 meters), the calculation is:

amax = 2764虏 rad虏/s虏 脳 0.00075 m = 5728.8 m/s虏.

This tells us that the diaphragm experiences a maximum acceleration of 5728.8 meters per second squared, which is a measure of how quickly the speed changes at the extremes of its motion.

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Most popular questions from this chapter

6\. Spring Balance The scale of a spring balance that reads from 0 to \(15.0 \mathrm{~kg}\) is \(12.0 \mathrm{~cm}\) long. A package suspended from the balance is found to oscillate vertically with a frequency of \(2.00 \mathrm{~Hz} .\) (a) What is the spring constant? (b) How much does the package weigh?

69\. To What Angle? A small metal ball of mass \(m\) hangs from a pivot by a rigid, light metal rod of length \(R .\) The ball is swinging back and forth with an amplitude that remains small throughout its motion, \(\theta^{\max } \leq 5^{\circ} .\) Ignore all damping. (a) The equation of motion of this ideal pendulum can be derived in a variety of ways and is $$ \frac{d^{2} \theta}{d t^{2}}=-\frac{g}{R} \sin \theta $$ For small angles, show how this can be replaced by an approximate equation of motion that can be solved more easily than the one given. (b) Write a general solution for the approximate equation of motion you obtained in (a) that works for any starting angle and angular velocity (as long as the angles stay in the range where the approximation is OK). Demonstrate that what you have written is a solution and show that at a time \(t=0\) your solution can have any given starting position and velocity. (c) If the length of the rod is \(0.3 \mathrm{~m}\), the mass of the ball is \(0.2 \mathrm{~kg}\), and the clock is started at a time when the ball is passing through the center \((\theta=0)\) and is moving with an angular speed of \(0.1 \mathrm{rad} / \mathrm{s}\), find the maximum angle your solution says the ball will reach. Can you use the approximate equation of motion for this motion? If the starting angle is not small, you cannot easily solve the equation of motion without a computer. But there are still things you can do. (d) Derive the energy conservation equation for the motion of the pendulum. (Do not use the small-amplitude approximation.) (e) If the pendulum is released from a starting angle of \(\theta_{1}\), what will be the maximum speed it travels at any point on its swing?

54\. Particle Undergoing SHM A \(10 \mathrm{~g}\) particle is undergoing simple harmonic motion with an amplitude of \(2.0 \times 10^{-3} \mathrm{~m}\) and a maximum acceleration of magnitude \(8.0 \times 10^{-3} \mathrm{~m} / \mathrm{s}^{2}\). The phase constant is \(-\pi / 3\) rad. (a) Write an equation for the force on the particle as a function of time. (b) What is the period of the motion? (c) What is the maximum speed of the particle? (d) What is the total mechanical energy of this simple harmonic oscillator?

16\. Two Blocks In Fig. 16-32 two blocks \((m=1.0 \mathrm{~kg}\) and \(M=\) \(10 \mathrm{~kg}\) ) and a spring \((k=200 \mathrm{~N} / \mathrm{m})\) are arranged on a horizontal, frictionless surface. The coefficient of static friction between the two blocks is \(0.40\). What amplitude of simple harmonic motion of the spring-blocks system puts the smaller block on the verge of slipping over the larger block?

2\. Oscillating Block An oscillating block-spring system takes \(0.75 \mathrm{~s}\) to begin repeating its motion. Find its (a) period, (b) frequency in hertz, and (c) angular frequency in radians per second.
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