5. Loudspeaker A loudspeaker produces a musical sound by means of the oscillation of a diaphragm

5. Loudspeaker A loudspeaker produces a musical sound by...

Key Concepts

Amplitude Limitation Impact

The amplitude of oscillation, although small, plays a significant role in the dynamics of the system since acceleration in SHM is proportional to both the square of the angular frequency and the amplitude. The limited amplitude directly influences the minimum frequency required to achieve accelerations above the given threshold.

Threshold Acceleration Condition

Setting a threshold acceleration condition, such as when the acceleration exceeds gravitational acceleration (g), provides a criterion to determine the critical frequencies of oscillation. By equating the maximum acceleration expression (?²A) to g, one can solve for the frequency values where the acceleration exceeds a specified limit.

Simple Harmonic Motion

Simple harmonic motion (SHM) describes a type of periodic oscillatory motion in which the restoring force is directly proportional to the displacement and acts in the opposite direction. This concept is central to understanding how systems like a loudspeaker diaphragm oscillate in a predictable, sinusoidal pattern.

Acceleration in SHM

In simple harmonic motion, the acceleration of an oscillating object is given by the relation a = ?²A, where ? is the angular frequency and A is the amplitude of oscillation. This implies that even with small amplitude oscillations, high frequencies (or large angular frequencies) can lead to significant accelerations.

Frequency and Angular Frequency Relationship

The relationship between frequency (f) and angular frequency (?) is described by the equation ? = 2?f. This conversion is critical when calculating acceleration, as the acceleration in SHM is directly dependent on the square of the angular frequency.

Transcript

00:01 So in this question we're asked to determine the frequencies for which the resulting magnitude of the acceleration of a diaphragm on a speaker exceeds g, given that the amplitude of oscillation is 1 by 10 to the minus 3 millimeters.

00:17 So let's go, let's determine what this question then.

00:22 So we're given that the amplitude of oscillation is 1 by 10 to the minus 3 millimeters, which is then equal to, if we work this out, this is equal to 1 by 10 to the minus 6 meters.

00:34 And we know what g is.

00:35 G is we're going to take it's going to be 9 .81 meters per second squared.

00:39 And we know that the acceleration of a body undergoing simple harmonic motion, which we're going to assume simple harmonic motion, is equal to omega squared times x of t.

00:48 So x is a function of t, which is equal the omega squared times the amplitude times cos omega t.

00:54 And we know that cost omega t is maximum for when it is equal to 1.

00:59 But if we're talking about magnitude, of acceleration it's one so one or minus one for maximum since we're taking the absolute value of these so that means that the maximum acceleration is given by a max is equal to omega square it times a so let's go by answering this question so let's find the value the frequency that the that the oscillator would have the diaphragm would have to oscillate at if we want the acceleration to be to be equal to so if we rearrange this, we get that omega is equal to the square root of a max, which you want to be g in this case over a, which is 1 by 10 to minus 6.

01:47 So if we work this out, this is equal to 3 .132 by 10 to the 3 rats per second.

01:59 And then if we work the frequency out, we got the frequency is equal to omega over 2 .5...

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