Respuesta :
Answer:
0.07756 m
Explanation:
Given mass of object =0.20 kg
spring constant = 120 n/m
maximum speed = 1.9 m/sec
We have to find the amplitude of the motion
We know that maximum speed of the object when it is in harmonic motion is given by [tex]v_{max}=A\omega[/tex] where A is amplitude and [tex]\omega[/tex] is angular velocity
Angular velocity is given by [tex]\omega=\sqrt{\frac{k}{m}}[/tex] where k is spring constant and m is mass
So [tex]v_{max}=A\sqrt{\frac{k}{m}}[/tex]
[tex]A=V_{max}\sqrt{\frac{m}{k}}=1.9\times \sqrt{\frac{0.2}{120}}=0.07756 \ m[/tex]
Answer:
The amplitude of the motion is 0.077 m
Explanation:
It is given that,
Mass of the object, m = 0.2 kg
Spring constant, K = 120 N/m
Maximum speed of the object, v = 1.9 m/s
We know that, in harmonic motion, the maximum speed of the object is given by :
[tex]v_{max}=A\omega[/tex]
Where
A is the amplitude of the wave
[tex]\omega[/tex] is the angular velocity, [tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
[tex]v_{max}=A\sqrt{\dfrac{k}{m}}[/tex]
[tex]A=v_{max}\sqrt{\dfrac{m}{K}}[/tex]
[tex]A=1.9\times \sqrt{\dfrac{0.2}{120}}[/tex]
A = 0.077 m
So, the amplitude of the motion is 0.077 m. Hence, this is the required solution.
