The sound pressure P for a given sound is given by: P= 10 log W/ Wo
Its units are decibels (dB). W is the size of a variable energy source (called the sound power), measured in Watts. Wo is the lowest threshold of sound that humans can typically hear. It is a constant given by:
Wo=10^-12 W/m2
a) Find the rate of change of the sound pressure P with respect to time if W=7.2 and dW/dt = 0.5 at some given time t.
b) If the variable sound power W is given by W = t2 +t +1, find the rate of change of the sound pressure P, at time t = 3s.
c) If W = cos 0.2t, find the rate of change of the sound pressure P, at time t = 15 (calculators in radians)

a. 0.302 dB/s

b. 2.34 dB/s

c. 1.24 dB/s

Step-by-step explanation:

a) Find the rate of change of the sound pressure P with respect to time if W=7.2 and dW/dt = 0.5 at some given time t.

Since P = 10log(W/W₀), its rate of change with respect to time is

dP/dt = dP/dW × dW/dt

dP/dW = d[10logW - 10logW₀]/dt

dP/dW = d[10lnW/2.303 - 10logW₀]/dt

= 10/2.303W - 0

= 10/(2.303W)

dP/dt = dP/dW × dW/dt

dP/dt = 10/(2.303W) × dW/dt

Since dW/dt = 0.5 when W = 7.2, then

dP/dt = 10/(2.303 × 7.2) × 0.5

dP/dt = 10/16.5816 × 0.5

dP/dt = 0.603 × 0.5

dP/dt = 0.302 dB/s

b) If the variable sound power W is given by W = t² + t + 1, find the rate of change of the sound pressure P, at time t = 3s.

Since P = 10log(W/W₀), its rate of change with respect to time is

dP/dt = dP/dW × dW/dt

dP/dW = d[10logW - 10logW₀]/dt

dP/dW = d[10lnW/2.303 - 10logW₀]/dt

= 10/2.303W - 0

= 10/(2.303W)

and dW/dt = d(t² + t + 1)/dt = 2t + 1

So, dP/dt = dP/dW × dW/dt

dP/dt = 10/(2.303W) × (2t + 1)

dP/dt = 10(2t + 1)/(2.303W)

dP/dt = 10(2t + 1)/[2.303(t² + t + 1)]

we then substitute t = 3 into the equation

dP/dt = 10(2t + 1)/[2.303(t² + t + 1)]

dP/dt = 10(2(3) + 1)/[2.303((3)² + 3 + 1)]

dP/dt = 10(6 + 1)/[2.303(9 + 3 + 1)]

dP/dt = 10(7)/[2.303(13)]

dP/dt = 70/29.939

dP/dt = 2.34 dB/s

c) If W = cos 0.2t, find the rate of change of the sound pressure P, at time t = 15 (calculators in radians)

Since dP/dt = dP/dW × dW/dt and dP/dW = 10/(2.303W) and W = cos(0.2t), dW/dt = -0.2sin(0.2t)

So, dP/dt = dP/dW × dW/dt

dP/dt = 10/(2.303W) × -0.2sin(0.2t)

dP/dt = -20sin(0.2t)/(2.303W)

dP/dt = -20sin(0.2t)/(2.303cos(0.2t))

dP/dt = -20tan(0.2t)/2.303

when t = 15, we have

dP/dt = -20tan(0.2 × 15)/2.303

dP/dt = -20tan3/2.303

dP/dt = -20 × -0.1425/2.303

dP/dt = 2.8509/2.303

dP/dt = 1.238

dP/dt ≅ 1.24 dB/s

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

i need a diagram

step-by-step explanation:

no