What is the pressure amplitude for the threshold of human hearing?
2 answers:
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Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so = A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Answer:
a.Beth
b.2232 s
Explanation:
We are given that
Distance,d=400 mi
Speed of Alan,v=45 mph
Speed of Beth,v'=55 mph
a.Time =
Using the formula
Time taken by Alan=
Time taken by Beth=
Alan will reach San Francisco at 4:53 PM
Beth will reach San Francisco at 4:16 PM
Beth will reach before Alan.
b.Difference between time=8.89-7.27=1.62 hr
t=1.62 hr
1.62-1=0.62 hr
0.62 hr=
Hence, Beth has to wait 2232 s for Alan to arrive .