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A body of mass 2kg suspended through a vertical spring executes SHM of period 4s. If the oscillation are stopped and body hangs in equilibrium,find the potential energy stored in the spring.?
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A body of mass 2kg suspended through a vertical spring executes SHM of...
Explanation:
The period of oscillation of a spring system can be given by the formula:
T=2π√(m/k)
where T is the period of oscillation, m is the mass of the object attached to the spring, and k is the spring constant.
In this problem, the period of oscillation is given as 4s and the mass of the object attached to the spring is 2kg. Therefore, we can write:
4=2π√(2/k)
Squaring both sides, we get:
16=4π^2(2/k)
Simplifying, we get:
k=π^2
Therefore, the spring constant is π^2 N/m.
Potential Energy Stored in the Spring:
When an object is attached to a spring and is displaced from its equilibrium position, the spring exerts a restoring force on the object that is proportional to the displacement. This restoring force stores potential energy in the spring.
The potential energy stored in a spring can be given by the formula:
U=1/2 k x^2
where U is the potential energy stored in the spring, k is the spring constant, and x is the displacement of the object from its equilibrium position.
In this problem, we are given that the oscillations are stopped and the body hangs in equilibrium. This means that the displacement of the object from its equilibrium position is zero. Therefore, the potential energy stored in the spring is also zero.
Conclusion:
In this problem, we have calculated the spring constant of a spring system and determined that the potential energy stored in the spring is zero when the object attached to the spring is in equilibrium.
The period of oscillation of a spring system can be given by the formula:
T=2π√(m/k)
where T is the period of oscillation, m is the mass of the object attached to the spring, and k is the spring constant.
In this problem, the period of oscillation is given as 4s and the mass of the object attached to the spring is 2kg. Therefore, we can write:
4=2π√(2/k)
Squaring both sides, we get:
16=4π^2(2/k)
Simplifying, we get:
k=π^2
Therefore, the spring constant is π^2 N/m.
Potential Energy Stored in the Spring:
When an object is attached to a spring and is displaced from its equilibrium position, the spring exerts a restoring force on the object that is proportional to the displacement. This restoring force stores potential energy in the spring.
The potential energy stored in a spring can be given by the formula:
U=1/2 k x^2
where U is the potential energy stored in the spring, k is the spring constant, and x is the displacement of the object from its equilibrium position.
In this problem, we are given that the oscillations are stopped and the body hangs in equilibrium. This means that the displacement of the object from its equilibrium position is zero. Therefore, the potential energy stored in the spring is also zero.
Conclusion:
In this problem, we have calculated the spring constant of a spring system and determined that the potential energy stored in the spring is zero when the object attached to the spring is in equilibrium.
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A body of mass 2kg suspended through a vertical spring executes SHM of...
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A body of mass 2kg suspended through a vertical spring executes SHM of period 4s. If the oscillation are stopped and body hangs in equilibrium,find the potential energy stored in the spring.?
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A body of mass 2kg suspended through a vertical spring executes SHM of period 4s. If the oscillation are stopped and body hangs in equilibrium,find the potential energy stored in the spring.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A body of mass 2kg suspended through a vertical spring executes SHM of period 4s. If the oscillation are stopped and body hangs in equilibrium,find the potential energy stored in the spring.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body of mass 2kg suspended through a vertical spring executes SHM of period 4s. If the oscillation are stopped and body hangs in equilibrium,find the potential energy stored in the spring.?.
A body of mass 2kg suspended through a vertical spring executes SHM of period 4s. If the oscillation are stopped and body hangs in equilibrium,find the potential energy stored in the spring.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A body of mass 2kg suspended through a vertical spring executes SHM of period 4s. If the oscillation are stopped and body hangs in equilibrium,find the potential energy stored in the spring.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body of mass 2kg suspended through a vertical spring executes SHM of period 4s. If the oscillation are stopped and body hangs in equilibrium,find the potential energy stored in the spring.?.
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