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Answer :
The block's oscillation frequency is [tex]\( 8 \, \text{Hz} \).[/tex]
To find the oscillation frequency of the block-spring system, we use the formula for the frequency of simple harmonic motion: [tex]\( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \),[/tex] where [tex]\( f \) is the frequency, \( k \)[/tex] is the force constant of the spring, and [tex]\( m \)[/tex] is the mass of the block.
First, we identify the given values:
[tex]\( k = 128 \, \text{N/m} \) and \( m = 0.720 \, \text{kg} \).[/tex]
Substituting these into the formula,
we get [tex]\( f = \frac{1}{2\pi} \sqrt{\frac{128 \, \text{N/m}}{0.720 \, \text{kg}}} \).[/tex]
Solving this yields [tex]\( f \approx 8 \, \text{Hz} \),[/tex] indicating that the block oscillates back and forth with a frequency of approximately [tex]\(8 \, \text{cycles per second}\).[/tex]
To find the oscillation frequency, we use the formula [tex]\( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \),[/tex]
where [tex]\( f \) is the frequency, \( k \)[/tex] is the force constant, and [tex]\( m \)[/tex] is the mass. Plugging in the values, we get [tex]\( f = \frac{1}{2\pi} \sqrt{\frac{128 \, \text{N/m}}{0.720 \, \text{kg}}} \).[/tex]
Calculating, we get [tex]\( f \approx 8 \, \text{Hz} \).[/tex]
This frequency represents the number of oscillations per second.
Therefore, the block oscillates back and forth with a frequency of [tex]\(8 \, \text{Hz}\).[/tex]
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