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A spring with [tex]$k = 33.5 \, \text{N/m}$[/tex] has a [tex]1.20 \, \text{kg}[/tex] mass attached. It is pulled [tex]0.120 \, \text{m}[/tex] and released.

What is its maximum speed?

(Unit [tex]= \text{m/s}[/tex])


Sagot :

To determine the maximum speed of a mass attached to a spring that has been displaced and released, we can use the formula for the maximum speed in simple harmonic motion:

[tex]\[ v_{\text{max}} = \sqrt{\frac{k}{m}} \cdot A \][/tex]

where:
- [tex]\( v_{\text{max}} \)[/tex] is the maximum speed,
- [tex]\( k \)[/tex] is the spring constant,
- [tex]\( m \)[/tex] is the mass,
- [tex]\( A \)[/tex] is the amplitude of the motion (which in this case is the initial displacement).

Given the data:
- Spring constant, [tex]\( k = 33.5 \, \text{N/m} \)[/tex]
- Mass, [tex]\( m = 1.20 \, \text{kg} \)[/tex]
- Displacement (amplitude), [tex]\( A = 0.120 \, \text{m} \)[/tex]

Let's go through the steps to calculate the maximum speed:

1. First, determine the ratio of the spring constant to the mass:
[tex]\[ \frac{k}{m} = \frac{33.5 \, \text{N/m}}{1.20 \, \text{kg}} \][/tex]

2. After calculating the ratio, take the square root of this ratio to get the angular frequency:
[tex]\[ \sqrt{\frac{k}{m}} \][/tex]

3. Multiply this result by the amplitude [tex]\( A \)[/tex]:
[tex]\[ v_{\text{max}} = \sqrt{\frac{33.5}{1.20}} \cdot 0.120 \][/tex]

Following these steps and using the correct values, the maximum speed comes out to be:
[tex]\[ v_{\text{max}} \approx 0.634 \, \text{m/s} \][/tex]

So, the maximum speed of the mass is approximately [tex]\( 0.634 \, \text{m/s} \)[/tex].