A 1.54 x 10-2-kg block is resting on a horizontal friction-less surface and is attached to a horizontal spring whose spring constant is 129 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 8.84 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?
v=Aw
A=v/w
w=sqrt of k/m
w=91.52
A=8.84/91.52
A=0.0966m
initial KE=1/2*m v^2
ending PE=1/2 k A^2
set them equal
1/2 k A^2=1/2 m v^2
A^2=mass/k v^2
= 1.54E-2/129 * 8.84^2 == 0.00932900961
A=sqrt (0.00932900961)
A== 0.0966 meters
To find the amplitude of the resulting simple harmonic motion, we need to determine the maximum displacement from equilibrium caused by the initial speed of the block.
In this case, we can use the principle of conservation of mechanical energy. The initial kinetic energy of the block will be converted into potential energy stored in the spring when it comes to rest at the maximum displacement.
The initial kinetic energy (K) of the block can be calculated using the formula:
K = (1/2)mv^2
where m is the mass (1.54 x 10^(-2) kg) and v is the initial speed (8.84 m/s).
Next, we determine the potential energy stored in the spring when the block comes to rest at the maximum displacement. The potential energy (U) stored in a spring can be calculated using the formula:
U = (1/2)kx^2
where k is the spring constant (129 N/m) and x is the maximum displacement (amplitude).
According to the conservation of mechanical energy principle, the initial kinetic energy is equal to the potential energy at the maximum displacement, so we can equate the two equations:
(1/2)mv^2 = (1/2)kx^2
Solving for x, we get:
x^2 = (mv^2) / k
Finally, taking the square root of both sides, we can find the amplitude (maximum displacement):
x = √[(mv^2) / k]
Substituting the given values, we get:
x = √[(1.54 x 10^(-2) kg * (8.84 m/s)^2) / 129 N/m]
By calculating this equation, we can determine the amplitude of the resulting simple harmonic motion.