Respuesta :
Explanation:
Given that,
Force constant of the spring, k = 120 N/m
Frequency of vibration, f = 6 Hz
Solution,
(a) Let T is the time period of the spring. The relation between the frequency and the time period is given by :
[tex]T=\dfrac{1}{f}[/tex]
[tex]T=\dfrac{1}{6}[/tex]
T = 0.167 s
(b) Let [tex]\omega[/tex] is the angular frequency. It is given by :
[tex]\omega=2\pi f[/tex]
[tex]\omega=2\pi \times 6[/tex]
[tex]\omega=37.69\ rad[/tex]
(c) The relation between angular frequency and the spring constant is given by :
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
[tex]m=\dfrac{k}{\omega^2}[/tex]
[tex]m=\dfrac{120}{(37.69)^2}[/tex]
m = 0.084 kg
or m = 84 grams
Therefore, it is the required solution.
Part(a): The period of vibration is [tex]0.167 s[/tex]
Part(b): The angular frequency of the vibration is [tex]37.68 rad/s[/tex]
Part(c): The mass of the body is 0.084 kg
Frequency of vibration:
The number of vibrations or oscillations made in one second is called the frequency of vibration.
Part(a):
Substitute the given value 6.00Hz in the equation [tex]T=\frac{1}{f}[/tex] we get,
[tex]T=\frac{1}{f} \\T=\frac{1}{6.00} \\T=0.167 s[/tex]
Part(b):
Substitute the given value 6.0Hz into the equation [tex]\omega =2\pi f[/tex]
[tex]\omega =2\pi(6.00)\\\omega=37.68 rad/sec[/tex]
Part(c):
Substitute the given value 12 N/m into the equation [tex]m=\frac{k}{\omega^2}[/tex] we get,
[tex]m=\frac{k}{\omega^2} \\m=\frac{120}{37.68} \\m=0.084 kg[/tex]
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