Answer:
0.075 m = 7.5 cm
Explanation:
In a simple harmonic motion, the frequency of oscillation (f) is related to the mass (m) and the spring constant (k) by the formula
[tex]f=\frac{1}{2\pi} \sqrt{\frac{k}{m}}[/tex]
In this problem, we know
f = 1.1 Hz
m = 0.28 kg
So we can re-arrange this formula to find the spring constant:
[tex]k=(2\pi f)^2 m=(2 \pi (1.1 Hz))^2 (0.28 kg)=13.4 N/m[/tex]
The restoring force of the spring is:
[tex]F=kx[/tex]
where
F = 1.0 N is the force exerted on the block
x is the displacement of the block
Therefore, by re-arranging this equation we can find how far was the block pulled back before being released:
[tex]x=\frac{F}{k}=\frac{1.0 N}{13.4 N/m}=0.075 m=7.5 cm[/tex]
The block was pulled at a distance of 0.0823m before being released
First, we need to get the spring constant using the formula for the frequency of the spring as shown;
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m} }[/tex]
Given the following parameters
mass m = 0.28kg
F = 1.0N
f = 1.1Hz
Substitute the given parameters into the formula to have:
[tex]1.1 = \frac{1}{6.28}\sqrt{\frac{k}{0.28} }\\6.908 = \sqrt{\frac{k}{0.28} }\\(6.908)^2=\frac{k}{0.28}\\k=0.28\times 47.72\\k= 13.3616N/m[/tex]
Next is to get the distance that the block is being pulled before release. According to Hooke's law;
F = kx
x = F/k
x = 1.1/13.3616
x = 0.0823m
Hence the block was pulled at a distance of 0.0823m before being released
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