# Block hanging from spring when a second block is added

• dtesselstrom
In summary, the problem involves a block hanging from a vertical spring in equilibrium. When a second identical block is added, the original block sags by 3.00 cm. The question is asking for the oscillation frequency of this system. The attempted solution involves solving for the spring constant using the formula F=-ky and then plugging it into T=2pi sq root of m/k, but this was deemed incorrect. Additional information on how to approach the problem would be appreciated. It is also noted that the 3cm sag is due to adding 1m and the spring's stretch from its natural length with 2m hanging on it is unknown.
dtesselstrom
A block hangs in equilibrium from a vertical spring. When a second identical block is added, the original block sags by 3.00 cm. It wants to know the oscillation frequency of this problem.
So far I tried solving for F=-ky with F being 2m*g and y equal to .03. I solved for k and plugged that into T=2pi sq root of m/k this however said it was wrong. Any information on how to approach this would be nice.

dtesselstrom said:
A block hangs in equilibrium from a vertical spring. When a second identical block is added, the original block sags by 3.00 cm. It wants to know the oscillation frequency of this problem.
So far I tried solving for F=-ky with F being 2m*g and y equal to .03. I solved for k and plugged that into T=2pi sq root of m/k this however said it was wrong. Any information on how to approach this would be nice.
The 3cm sag comes from adding 1m. How much do you think the spring is streteched from its natural length with the 2m hanging on it?

Based on the given information, it seems like the second block causes the original block to sag by 3.00 cm, indicating that the added weight has affected the equilibrium position of the system. This means that the spring constant (k) of the spring has also changed.

To accurately calculate the oscillation frequency of the system, we need to take into account the new equilibrium position and the new spring constant. The correct equation to use would be T=2pi sq root of m/(k+2k), where m is the mass of one block and k is the original spring constant.

In order to solve for k, we can use the given information that the original block sags by 3.00 cm. This means that the new equilibrium position of the system is 3.00 cm below the original equilibrium position. We can use this to set up the equation kx=mg, where x is the displacement from the original equilibrium position and m is the mass of one block.

Solving for k, we get k=mg/x. Plugging this into the equation for T, we get T=2pi sq root of m/((mg/x)+2(mg/x)) = 2pi sq root of m/(3mg/x) = 2pi sq root of 3x/g.

Therefore, the oscillation frequency of the system would be 2pi sq root of 3x/g, where x is the displacement from the original equilibrium position (in this case, 3.00 cm).

I hope this helps you in solving the problem. It is important to carefully consider all the given information and use the appropriate equations to accurately calculate the desired result.

## 1. How does adding a second block affect the hanging block's motion?

Adding a second block to the hanging block-spring system will increase the total mass of the system, causing the frequency of the motion to decrease. This means that the hanging block will take longer to complete one full cycle.

## 2. Will the amplitude of the hanging block's motion change with the addition of a second block?

Assuming all other variables remain constant, the amplitude of the hanging block's motion will not change with the addition of a second block. The amplitude is determined by the initial displacement of the hanging block and the spring constant.

## 3. How does the spring constant affect the motion of the hanging block when a second block is added?

The spring constant determines the stiffness of the spring and is directly proportional to the frequency of the motion. Therefore, when a second block is added, the spring constant will affect how quickly the hanging block completes one full cycle of motion.

## 4. What is the relationship between the mass of the second block and the motion of the hanging block?

The mass of the second block will affect the frequency of the motion, with a heavier mass causing a decrease in frequency. This is because the added mass will require more force to move and will take longer to complete one full cycle of motion.

## 5. Can the motion of the hanging block be affected by changing the position of the second block?

Yes, the position of the second block can affect the motion of the hanging block. Placing the second block closer to the hanging block will increase the total mass of the system, causing a decrease in frequency. Placing the second block further away will decrease the total mass and increase the frequency of the motion.

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