A 8.50-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical
oscillations having a period of 1.70 s. Find the force constant of the spring
2 answers:
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Answer:
The intensity level in the room is 63 dB
Explanation:
To calculate the intensity of sound in the room, we use the equation of definition of decibels
β = 10 log (I / Io) (1)
With “I” the sound intensity and “Io” the threshold intensity 1.0 10⁻⁻¹² W/m²
To calculate the intensity we will use the initial data and remember the power of the emitted sound is constant, in addition that the sound propagates in three-dimensional form or on a spherical surface
I = P/A ⇒ P = I A
The area of a sphere is 4 π r², where I can calculate of 1
β/10 = log (I/Io)
I / Io =
I = Io
I = 1 10⁻¹² 10⁽¹⁰⁰/¹⁰⁾ = 1 10⁻¹² 10¹⁰
I = 1.0 10⁻² W
With this we can calculate the intensity for a distance of 20 m
I = 1.0 10⁻² / ( 4π 20²)
I = 2.0 10⁻⁶ W/m²
We have already found the intensity at the point of interest, so we can calculate the intensity in decibels at this point with equation 1
β = 10 log(2.0 10⁻⁶ / 1.0 10⁻¹²)
β = 10 log ( 2 10⁶) = 10 6.3
β = 63 dB
The intensity level in the room is 63 dB