Question

A loudspeaker diaphragm is producing a sound by moving back and forth in a simple harmonic motion. the angular frequency of the motion is 8.02×104 rad s-1 . how many times does the diaphragm move back and forth in 1 minute?

Answer 1

The diaphragm moves back and forth approximately
`\(7.656 * 10^5\)` times in 1 minute.

To find the number of oscillations or cycles n in 1 minute, we can use the relationship between angular frequency
`(\(\omega\))` and frequency f in simple harmonic motion:

`\[ \omega = 2\pi f \]`

Thus:

`\[ f = (\omega)/(2\pi) \]`

Given the angular frequency
`(\(\omega\))` as
`\(8.02 * 10^4 \, \text{rad/s}\)`, we can substitute this into the formula:

`\[ f = (8.02 * 10^4)/(2\pi) \]`

Now, to find the number of cycles (\(n\)) in 1 minute, we can use the formula:

`\[ n = f * \text{time in seconds} \]`

Since there are 60 seconds in 1 minute, we substitute \(60\) seconds for the time:

`\[ n = f * 60 \]`

Calculate Frequency f:

`\[ f = (\omega)/(2\pi) = (8.02 * 10^4)/(2\pi) \]`

`\[ f \approx (8.02 * 10^4)/(2 * 3.1416) \]`

`\[ f \approx (8.02 * 10^4)/(6.2832) \]`

`\[ f \approx 1.276 * 10^4 \, \text{Hz} \]`

Number of Cycles n in 1 Minute:
`\[ n = f * \text{time in seconds} \]`

Since there are 60 seconds in 1 minute:

`\[ n = 1.276 * 10^4 * 60 \]`

`\[ n \approx 7.656 * 10^5 \, \text{cycles} \]`

Therefore, the diaphragm moves back and forth approximately
`\(7.656 * 10^5\)` times in 1 minute.