A block of mass m=.75kg is fastened to an unstrained horizontal springwhose spring constant is 82 N/m. The block is given a displacement of +.12m, where + sign indicates that the displacement is along the +x axis, and then released from rest.
Find the angular frequency w of the resulting oscillatory motion.
For this question, I got 10.646. Is this correct?
What is the maximum speed of the block?
Would I need to find the amplitude to multiply by w to get the maximum speed? If that's the case, how would I find the amplitude?
Thank you!
The angular frequency w is sqrt (k/m). sqrt (82/.75) = 10.456 sec^-1
The maximum speed V can be obtained from the stored potential energy is the spring before release.
(1/2) k X^2 = (1/2) m V^2
V = sqrt (k/m) X = w X
where X is the 0.12 m displacement
To find the angular frequency (ω) of the resulting oscillatory motion, you can use the formula:
ω = sqrt(k / m),
where k is the spring constant and m is the mass of the block.
Given that the spring constant is 82 N/m and the mass is 0.75 kg, you can calculate the angular frequency as follows:
ω = sqrt(82 N/m / 0.75 kg)
= sqrt(109.33 N/kg)
≈ 10.46 rad/s
So, your answer of approximately 10.646 rad/s is correct.
To find the maximum speed of the block in the oscillatory motion, you do not need to find the amplitude. The maximum speed of the block occurs when it passes through the equilibrium position (displacement = 0). At this point, all of the potential energy is converted into kinetic energy.
The total mechanical energy (E) of the oscillating system is given by:
E = (1/2)kA^2,
where A is the amplitude of the motion. Since the block is released from rest, the initial velocity is 0, and therefore, the initial kinetic energy is also 0. Hence, at any point during the motion, the total mechanical energy will be equal to the potential energy at that point, given by:
E = (1/2)kx^2,
where x is the displacement of the block from the equilibrium position.
When the block is at the maximum displacement (amplitude A), its potential energy is at the maximum, given by:
E = (1/2)kA^2.
At this point, all of the potential energy is converted into kinetic energy, so:
E = (1/2)mv^2,
where v is the maximum speed of the block.
Setting the two equations equal to each other, we have:
(1/2)kA^2 = (1/2)mv^2.
Since the amplitude A is given as 0.12 m, and you have already calculated the angular frequency ω, you can use the formula:
v = Aω,
to find the maximum speed of the block:
v = 0.12 m * 10.46 rad/s
≈ 1.25 m/s.
Therefore, the maximum speed of the block is approximately 1.25 m/s.
Note: Amplitude (A) is given as the maximum displacement from the equilibrium position, so you can directly use it to find the maximum speed using the formula v = Aω, as explained above.