To solve this problem, we need to understand that when the block is at the extreme position, the entire energy of the system is potential energy stored in the spring.

The potential energy stored in a spring is given by the formula (1/2)kx², where k is the spring constant and x is the extension of the spring.

In this case, the total energy of the system is also equal to the kinetic energy of the block at the mean position, which is (1/2)mω²A², where m is the mass of the block, ω is the angular frequency, and A is the amplitude of oscillation.

Setting these two expressions equal to each other gives us:

(1/2)k1x² = (1/2)mω²A²

Solving for x (the extension of the spring) gives us:

x = sqrt((mω²A²)/k1)

This is the extension of the spring when the block is at the extreme position.

Question

To solve this problem, we need to understand that when the block is at the extreme position, the entire energy of the system is potential energy stored in the spring.

The potential energy stored in a spring is given by the formula (1/2)kx², where k is the spring constant and x is the extension of the spring.

In this case, the total energy of the system is also equal to the kinetic energy of the block at the mean position, which is (1/2)mω²A², where m is the mass of the block, ω is the angular frequency, and A is the amplitude of oscillation.

Setting these two expressions equal to each other gives us:

(1/2)k1x² = (1/2)mω²A²

Solving for x (the extension of the spring) gives us:

x = sqrt((mω²A²)/k1)

This is the extension of the spring when the block is at the extreme position.

Knowee AI · Accepted Answer

To solve this problem, we need to understand that when the block is at the extreme position, the entire energy of the system is potential energy stored in the spring.

The potential energy stored in a spring is given by the formula (1/2)kx², where k is the spring constant and x is the extension of the spring.

In this case, the total energy of the system is also equal to the kinetic energy of the block at the mean position, which is (1/2)mω²A², where m is the mass of the block, ω is the angular frequency, and A is the amplitude of oscillation.

Setting these two expressions equal to each other gives us:

(1/2)k1x² = (1/2)mω²A²

Solving for x (the extension of the spring) gives us:

x = sqrt((mω²A²)/k1)

This is the extension of the spring when the block is at the extreme position.

All the surface are smooth and thesystem, given below is oscillating withan amplitude A. What is the extensionof spring having spring constant 𝑘1,when the block is at the extremeposition?

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