Answer:
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
