SHM vertical spring problem

  • #1
torinketo
2
1
Homework Statement
A block of unknown mass is attached to the end of a vertical spring. When a second 50 g block is suspended, the spring extends by 38 cm. The oscillation period without the second 50 g block is 0.8 s. Find:

(a) the spring constant of the spring (in N/m);
(b) the mass of the first block (in kg).
Relevant Equations
T=2pi*sqrt(m/k)
k=mg/x
m1 = unknown
m2 = 0.05 kg
x = 0.38 m
T = 0.8s

Tried to plug in values into the above equations:

0.8 = 2pi*sqrt(m1+0.05/k)
k = ((m1+0.05)*9.8)/0.38

Got negative values for both k and m which doesn't make sense
 
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  • #2
torinketo said:
0.8 = 2pi*sqrt(m1+0.05/k)
The 0.8 seconds is for m1 by itself.
torinketo said:
k = ((m1+0.05)*9.8)/0.38
It's not clear whether the 0.38m is for the masses together or just the 50g mass.
 
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  • #3
haruspex said:
The 0.8 seconds is for m1 by itself.

It's not clear whether the 0.38m is for the masses together or just the 50g mass.
Thank you very much, this was the right answer :)

Screenshot_5.png
 

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  • #4
torinketo said:
Thank you very much, this was the right answer :)

View attachment 338824
Good, but you show too many significant figures. The data you were given only have two (the 0.8s only one, but probably intends 0.80s). And you should always state the units. So better answers are 1.3 kg/s2 and 0.021kg.
 
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1. What is SHM in a vertical spring problem?

SHM stands for Simple Harmonic Motion. In the context of a vertical spring problem, SHM refers to the oscillatory motion of a mass attached to a spring that is being stretched or compressed vertically.

2. How is the period of oscillation related to the spring constant in a vertical spring problem?

In a vertical spring problem, the period of oscillation is inversely proportional to the square root of the spring constant. This means that as the spring constant increases, the period of oscillation decreases.

3. What factors affect the amplitude of oscillation in a vertical spring problem?

The amplitude of oscillation in a vertical spring problem is affected by the initial displacement of the mass from its equilibrium position and the energy stored in the spring. A larger initial displacement or more energy stored in the spring will result in a larger amplitude of oscillation.

4. How does the mass of the object affect the frequency of oscillation in a vertical spring problem?

In a vertical spring problem, the frequency of oscillation is independent of the mass of the object attached to the spring. This means that the mass of the object does not affect how quickly the object oscillates up and down on the spring.

5. What is the equilibrium position in a vertical spring problem?

The equilibrium position in a vertical spring problem is the position where the spring is neither stretched nor compressed, and the object attached to the spring is at rest. It is the position where the net force on the object is zero.

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