During a typical workday (eight hours), the average sound intensity arriving at

Question

During a typical workday (eight hours), the average sound intensity arriving at Larry's ear is 1.8 times 10^-5 W/m. If the area of Larry's ear through which the sound passes is 2.1 times 10^-3 m^2, what is the total energy entering each of Larry's ears during the workday? The period of a simple pendulum in a grandfather clock on another planet is 1.66 s. What is the acceleration due to gravity on this planet? Assume that the length of the pendulum is 1.00 m. A mass m = 8.0 kg is attached to a spring and allowed to hang in the Earth's gravitational field. The spring stretches 3.6 cm from its equilibrium position. If allowed to oscillate, what would be its frequency? A performer seated on a trapeze is swinging back and forth with a period of 8.85 s. If she stands up, thus raising the center of mass of the trapeze + performer system by 35.0 cm, what will be the new period of the system? Treat trapeze + performer as a simple pendulum. The sound intensity level is reported in decibels. If one doubles the intensity of sound, by what f for does the perceived loudness, in decibels, change?

Explanation / Answer

6)

we know Intensity = power/Area

==> power = Intensity*Area

Energy = power*time

= Intensity*Area*time

= 1.8*10^-5*2.1*10^-3*24*60*60

= 3.26*10^-3 J

7)

we know, T = 2*pi*(L/g)

T^2 = 4*pi^2*(L/g)

==> g = 4*pi^2*L/T^2

= 4*pi^2*1/1.66^2

= 14.3 m/s^2

8)

from the given data,

spring constant, k =m*g/x

= 8*9.8/0.036

= 218 N/m

we know, angular freqquency, w = sqrt(k/m)

2*pi*f = sqrt(k/m)

f = sqrt(k/m)/(2*pi)

= sqrt(218/8)/(2*pi)

= 0.83 Hz

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