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A 2g block attached to an ideal spring with a spring constant of 80N/m oscillates on a horizantal friction less surface. When the spring is 4.0cm shorter than it's equilibrium length, the speed of the block is √17 m/s .The greatest speed of the block is?
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A 2g block attached to an ideal spring with a spring constant of 80N/m...
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A 2g block attached to an ideal spring with a spring constant of 80N/m...
Analysis:
The given situation involves a block attached to an ideal spring on a frictionless surface, undergoing simple harmonic motion.
Given Data:
- Mass of the block (m) = 2g = 0.002 kg
- Spring constant (k) = 80 N/m
- Compression of the spring (Δx) = 4.0 cm = 0.04 m
- Speed of the block at 4.0 cm compression = √17 m/s
Calculation:
Using the equation for the speed of the block in simple harmonic motion:
v = ω√(A^2 - x^2)
where:
- v is the speed of the block
- ω is the angular frequency of the oscillation
- A is the amplitude of the oscillation
- x is the displacement from equilibrium
Given that the block's speed at 4.0 cm compression is √17 m/s, we can calculate the amplitude of the oscillation:
√17 = ω√(0.04^2 - 0)
Solving for ω, we get:
ω = √17 / 0.04 = 2√17 rad/s
The maximum speed of the block occurs at the equilibrium position, where x = 0. Therefore, the maximum speed (v_max) is given by:
v_max = ωA
Since v_max is the greatest speed of the block, we substitute the values of ω and A into the equation:
v_max = 2√17 * A
Conclusion:
To find the greatest speed of the block, we need to determine the amplitude of the oscillation. By substituting the values into the equations and solving, we can calculate the maximum speed of the block in simple harmonic motion.
The given situation involves a block attached to an ideal spring on a frictionless surface, undergoing simple harmonic motion.
Given Data:
- Mass of the block (m) = 2g = 0.002 kg
- Spring constant (k) = 80 N/m
- Compression of the spring (Δx) = 4.0 cm = 0.04 m
- Speed of the block at 4.0 cm compression = √17 m/s
Calculation:
Using the equation for the speed of the block in simple harmonic motion:
v = ω√(A^2 - x^2)
where:
- v is the speed of the block
- ω is the angular frequency of the oscillation
- A is the amplitude of the oscillation
- x is the displacement from equilibrium
Given that the block's speed at 4.0 cm compression is √17 m/s, we can calculate the amplitude of the oscillation:
√17 = ω√(0.04^2 - 0)
Solving for ω, we get:
ω = √17 / 0.04 = 2√17 rad/s
The maximum speed of the block occurs at the equilibrium position, where x = 0. Therefore, the maximum speed (v_max) is given by:
v_max = ωA
Since v_max is the greatest speed of the block, we substitute the values of ω and A into the equation:
v_max = 2√17 * A
Conclusion:
To find the greatest speed of the block, we need to determine the amplitude of the oscillation. By substituting the values into the equations and solving, we can calculate the maximum speed of the block in simple harmonic motion.
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A 2g block attached to an ideal spring with a spring constant of 80N/m oscillates on a horizantal friction less surface. When the spring is 4.0cm shorter than it's equilibrium length, the speed of the block is √17 m/s .The greatest speed of the block is?
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A 2g block attached to an ideal spring with a spring constant of 80N/m oscillates on a horizantal friction less surface. When the spring is 4.0cm shorter than it's equilibrium length, the speed of the block is √17 m/s .The greatest speed of the block is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 2g block attached to an ideal spring with a spring constant of 80N/m oscillates on a horizantal friction less surface. When the spring is 4.0cm shorter than it's equilibrium length, the speed of the block is √17 m/s .The greatest speed of the block is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2g block attached to an ideal spring with a spring constant of 80N/m oscillates on a horizantal friction less surface. When the spring is 4.0cm shorter than it's equilibrium length, the speed of the block is √17 m/s .The greatest speed of the block is?.
A 2g block attached to an ideal spring with a spring constant of 80N/m oscillates on a horizantal friction less surface. When the spring is 4.0cm shorter than it's equilibrium length, the speed of the block is √17 m/s .The greatest speed of the block is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 2g block attached to an ideal spring with a spring constant of 80N/m oscillates on a horizantal friction less surface. When the spring is 4.0cm shorter than it's equilibrium length, the speed of the block is √17 m/s .The greatest speed of the block is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2g block attached to an ideal spring with a spring constant of 80N/m oscillates on a horizantal friction less surface. When the spring is 4.0cm shorter than it's equilibrium length, the speed of the block is √17 m/s .The greatest speed of the block is?.
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