Answer :
Answer:
Explanation:
mass attached m = .14 kg
force constant k = 5N / m
displacement
= amplitude of oscillation
A = .03 m
A ) period of motion = [tex]2\pi\sqrt{\frac{m}{k} }[/tex]
= 2 x 3.14 [tex]\sqrt{\frac{.14}{5} }[/tex]
T = 1.05 s
B ) maximum speed of block = angular velocity x amplitude
= (2π /T) x A
= (2 x 3.14 x .03) / 1.05
= .1794 m / s
17.94 cm /s
C )
maximum acceleration = angular velocity² x amplitude
= (2π /T)² x A
= (2π /1.05)² x .03
= 1.073 m / s²
D )
position
S = A cos ωt , ω is angular velocity
S = .03 cos(2πt /T)
= .03 cos 5.98 t
v =ω A sin(2πt /T)
= 5.98 x .03 sin5.98t
= .1794 sin5.98t
acceleration = ω²A sin5.98t
= 1.073 sin5.98t
For the mass connected to a light spring, the value of period, maximum speed and acceleration is,
- (A) The period of its motion is 1.05 seconds.
- (B) The maximum speed of the block is 0.1794 m/s.
- (C) The maximum acceleration of the block is 1.073 m/s ²
- (D) The position, velocity, and acceleration as functions of time in SI units is 0.3cos(5.98t) m, 0.1794sin(5.98t) and 1.073sin(5.98t) m/s² respectively.
What is angular speed of a body?
The angular speed of a body is the rate by which the body changed its angle with respect to the time. It can be given as,
[tex]\omega= \dfrac{\Delta \theta}{\Delta t}[/tex]
A 140 g mass is connected to a light spring of force constant 5 N/m that is free to oscillate on a horizontal, frictionless track.
The mass is displaced 3 cm from the equilibrium point and released from rest.
- (A) The period of its motion-
The period of motion is given by the following formula,
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
Here m, is the mass and k, is the spring constant. Mass of the spring is 140g or 0.14 kg, and spring constant is 5 N/m. Thus the period id,
[tex]T=2\pi\sqrt{\dfrac{0.14}{5}}\\T=1.05\rm s[/tex]
- (B) The maximum speed of the block-
First find out the angular velocity. Angular velocity can be given as,
[tex]\omega =\dfrac{2\pi}{T}\\\omega=\dfrac{2\pi}{1.05}\\\omega=5.98\rm rad/s[/tex]
Maximum speed of the block is product of angular velocity and the amplitude. The amplitude is 3 cm or 0.03 m. Therefore, speed is,
[tex]v=(\dfrac{2\pi}{1.05})^2\times0.3\\v=.1794m/s[/tex]
- (C) The maximum acceleration of the block-
Maximum acceleration of the block is product of square of angular velocity and the amplitude. The amplitude is 3 cm or 0.03 m. Therefore, acceleration is,
[tex]a=(\dfrac{2\pi}{1.05})^2\times0.3\\a=1.073\rm m/s^2[/tex]
(D) The position, velocity, and acceleration as functions of time in SI units.
The position can be given with cosine graph,
[tex]d=0.3\cos(\dfrac{(2\pi)}{1.05}t)\\d=0.3\cos(5.98t)\rm m[/tex]
The velocity can be given as,
[tex]v=v_{max}\sin(5.98t)\\v=0.1794\sin(5.98t)\rm m/s[/tex]
The acceleration is,
[tex]a=a_{max}\sin(5.98t)\\v=1.073\sin(5.98t)\rm m/s^2[/tex]
- (A) The period of its motion is 1.05 seconds.
- (B) The maximum speed of the block is 0.1794 m/s.
- (C) The maximum acceleration of the block is 1.073 m/s ²
- (D) The position, velocity, and acceleration as functions of time in SI units is 0.3cos(5.98t) m, 0.1794sin(5.98t) and 1.073sin(5.98t) m/s² respectively.
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