A spring of spring constant 30.0 N/m is attached to a 2.3 kg mass and set in motion. What is the period and frequency of vibrati
1 answer:
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Answer:
a) I = 3.63 W / m² , b) I = 0.750 W / m²
Explanation:
The intensity of a sound wave is given by the relation
I = P / A = ½ ρ v (2π f )²
I = (½ ρ v 4π² s_{max}²) f²
a) with the initial condition let's call the intensity Io
cte = (½ ρ v 4π² s_{max}²)
I₀ = cte s² f₀²
I₀ = cte 10 6
If frequency is increase f = 2.20 10³ Hz
I = constant (2.20 10³) 2
I = cte 4.84 10⁶
let's find the relationship of the two quantities
I / Io = 4.84
I = 4.84 Io
I = 4.84 0.750
I = 3.63 W / m²
b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or
I = cte (f s)²
I = constant (0.250 10³ 4)²
I = cte 1 10⁶
the relationship
I / Io = 1
I = Io
I = 0.750 W / m²
Answer:
t = 4 s
Explanation:
As we know that the particle A starts from Rest with constant acceleration
So the distance moved by the particle in given time "t"
Now we know that B moves with constant speed so in the same time B will move to another distance
now we know that B is already 349 cm down the track
so if A and B will meet after time "t"
then in that case
on solving above kinematics equation we have