Answer:
(a) 9.375 N/m
(b) 0.5024 m/s
(c) 0.01 kg/s
Explanation:
mass of head, m = 15 g = 0.015 kg
frequency, f = 4 Hz
Time period, T = 1 / f = 0.25 s
Let k is the spring constant.
(a)
The formula for the time period is
[tex]T=2\pi\sqrt{\frac{m}{K}}[/tex]
[tex]0.25=2\times 3.14 \sqrt{\frac{0.015}{K}}[/tex]
[tex]0.04=\sqrt{\frac{0.015}{K}}[/tex]
K = 9.375 N/m
(b)
Amplitude, A = 2 cm
Let ω is the angular velocity.
Maximum velocity, v = A ω = A x 2πf
v = 0.02 x 2 x 3.14 x 4 = 0.5024 m/s
(c)
Let b is the damping constant.
A(t = 4s) = 0.5 cm
Ao = 2 cm
Using the formula of damping
[tex]\frac{A}{A_{0}}=e^{-\frac{bt}{2m}}[/tex]
[tex]\frac{0.5}{2}}=e^{-\frac{b\times 4}{2\times 0.015}}[/tex]
[tex]0.25=e^{-133.3 b}[/tex]
Taking natural log on both the sides
ln (0.25) = - 133.3 b
- 133.3 b = - 1.386
b = 0.01 kg/s
This question involves the concepts of simple harmonic motion, spring constant, and amplitude.
a) The spring constant of the spring is "9.47 N/m".
b) The maximum speed of the head is "0.5 m/s".
c) The damping constant is "0.01 kg/s".
a)
We can find the spring constant of the spring by using the formula of frequency in the simple harmonic motion:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
where,
f = frequency = 4 Hz
k = spring constant = ?
m = mass = 15 g = 0.015 kg
Therefore,
[tex]4\ Hz=\frac{1}{2\pi}\sqrt{\frac{k}{0.015\ kg}}\\\\(16\ Hz^2)(4\pi^2)(0.015\ kg)=k\\[/tex]
k = 9.47 N/m
b)
Maximum speed is simply given by the following formula:
[tex]v=A\omega[/tex]
where,
v = maximum speed = ?
A = Amplitude = 2 cm = 0.02 m
ω = angular freuency = 2πf
Therefore,
[tex]v=A(2\pi f)=(0.02\ m)(2\pi)(4\ Hz)[/tex]
v = 0.5 m/s
c)
using the following equation to find out the damping constant:
[tex]ln(\frac{A}{A_o})=-\frac{bt}{2m}[/tex]
where,
A = amplitude at t = 4 s = 0.5 cm
A₀ = initial amplitude = 2 cm
b = damping constant = ?
t = time = 4 s
m = mass = 15 g = 0.015 kg
Therefore,
[tex]ln(\frac{0.5\ cm}{2\ cm})=-\frac{b(4\ s)}{2*0.015\ kg}[/tex]
[tex]\frac{(-1.386)(2)(0.015\ kg)}{4\ s}=-b[/tex]
b = 0.01 kg/s
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