Answer:
velocity changes over time.
Answer:
The frequencies are 
Explanation:
From the question we are told that
The length of the ear canal is 
The speed of sound is assumed to be 
Now taking look at a typical ear canal we see that we assume it is a closed pipe
Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

substituting values


Also the the second harmonic for the pipe (ear canal) is mathematically represented as
substituting values
Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies
Answer:
a) = 10.22 rad/s
b) = 0.35 m
Explanation:
Given
Mass of the particle, m = 1.1 kg
Force constant of the spring, k = 115 N/m
Distance at which the mass is released, d = 0.35 m
According to the differential equation of s Simple Harmonic Motion,
ω² = k / m, where
ω = angular frequency in rad/s
k = force constant in N/m
m = mass in kg
So,
ω² = 115 / 1.1
ω² = 104.55
ω = √104.55
ω = 10.22 rad/s
If y(0) = -0.35 m and we want our A to be positive, then suffice to say,
The value of coefficient A in meters is 0.35 m