JEE Exam > JEE Questions > A 0.5 kg block attached to an ideal spring wi...
Start Learning for Free
A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is?
Most Upvoted Answer
A 0.5 kg block attached to an ideal spring with spring constant 80 New...
This is SHM.
k = 80 N/m m = 0.50 kgω = angular frequency of SHM of spring mass system=> ω² = k/m = 160=> ω = 4√10 rad/sec²Let the equation of SHM be: x = x₀ Sin (4√10 t) and v = x₀ * 4√10 Cos( 4√10 t)
Given that displacement 0.04 meters = x₀ Sin(4√10 t)and speed: 0.50 m/s = x₀ 4√10 * Cos (4√10 t)
Hence, 0.04² + (0.50/4√10)² = x₀² x₀ = 0.0562 meters or 5.62 cm
Thus the maximum speed of the block = x₀ * ω = 0.711 m/sec.....
k = 80 N/m m = 0.50 kgω = angular frequency of SHM of spring mass system=> ω² = k/m = 160=> ω = 4√10 rad/sec²Let the equation of SHM be: x = x₀ Sin (4√10 t) and v = x₀ * 4√10 Cos( 4√10 t)
Given that displacement 0.04 meters = x₀ Sin(4√10 t)and speed: 0.50 m/s = x₀ 4√10 * Cos (4√10 t)
Hence, 0.04² + (0.50/4√10)² = x₀² x₀ = 0.0562 meters or 5.62 cm
Thus the maximum speed of the block = x₀ * ω = 0.711 m/sec.....
Community Answer
A 0.5 kg block attached to an ideal spring with spring constant 80 New...
Introduction:
The problem states that a 0.5 kg block is attached to an ideal spring with a spring constant of 80 N/m. The block oscillates on a horizontal frictionless surface, and when the spring is compressed by 4 cm from its equilibrium position, the block has a speed of 0.5 m/s. We need to find the greatest speed of the block during its oscillation.
Key Information:
- Mass of the block (m) = 0.5 kg
- Spring constant (k) = 80 N/m
- Displacement of the spring from equilibrium (x) = 4 cm = 0.04 m
- Speed of the block (v) = 0.5 m/s
Analysis:
To find the greatest speed of the block, we need to consider the conservation of mechanical energy. In simple harmonic motion, the total mechanical energy remains constant.
The total mechanical energy of the system is given by the sum of the potential energy stored in the spring and the kinetic energy of the block:
E = PE + KE
1. Potential Energy (PE):
The potential energy stored in the spring is given by the formula:
PE = (1/2)kx²
Substituting the given values, we can calculate the potential energy:
PE = (1/2) * 80 * (0.04)²
PE = 0.064 J
2. Kinetic Energy (KE):
The kinetic energy of the block is given by the formula:
KE = (1/2)mv²
Substituting the given values, we can calculate the kinetic energy:
0.5 = (1/2) * 0.5 * v²
v² = 1
v = 1 m/s
3. Total Mechanical Energy:
Since the total mechanical energy remains constant, we can equate the potential energy and kinetic energy:
PE = KE
0.064 = (1/2) * 0.5 * v²
0.064 = 0.25 * v²
Solving for v, we get:
v² = 0.256
v = √0.256
v ≈ 0.506 m/s
Conclusion:
Therefore, the greatest speed of the block during its oscillation is approximately 0.506 m/s. The block reaches this maximum speed when the spring is at its equilibrium position.
The problem states that a 0.5 kg block is attached to an ideal spring with a spring constant of 80 N/m. The block oscillates on a horizontal frictionless surface, and when the spring is compressed by 4 cm from its equilibrium position, the block has a speed of 0.5 m/s. We need to find the greatest speed of the block during its oscillation.
Key Information:
- Mass of the block (m) = 0.5 kg
- Spring constant (k) = 80 N/m
- Displacement of the spring from equilibrium (x) = 4 cm = 0.04 m
- Speed of the block (v) = 0.5 m/s
Analysis:
To find the greatest speed of the block, we need to consider the conservation of mechanical energy. In simple harmonic motion, the total mechanical energy remains constant.
The total mechanical energy of the system is given by the sum of the potential energy stored in the spring and the kinetic energy of the block:
E = PE + KE
1. Potential Energy (PE):
The potential energy stored in the spring is given by the formula:
PE = (1/2)kx²
Substituting the given values, we can calculate the potential energy:
PE = (1/2) * 80 * (0.04)²
PE = 0.064 J
2. Kinetic Energy (KE):
The kinetic energy of the block is given by the formula:
KE = (1/2)mv²
Substituting the given values, we can calculate the kinetic energy:
0.5 = (1/2) * 0.5 * v²
v² = 1
v = 1 m/s
3. Total Mechanical Energy:
Since the total mechanical energy remains constant, we can equate the potential energy and kinetic energy:
PE = KE
0.064 = (1/2) * 0.5 * v²
0.064 = 0.25 * v²
Solving for v, we get:
v² = 0.256
v = √0.256
v ≈ 0.506 m/s
Conclusion:
Therefore, the greatest speed of the block during its oscillation is approximately 0.506 m/s. The block reaches this maximum speed when the spring is at its equilibrium position.
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is?
Question Description
A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. 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 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. 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 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is?.
A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. 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 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. 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 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is?.
Solutions for A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is? defined & explained in the simplest way possible. Besides giving the explanation of
A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is?, a detailed solution for A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is? has been provided alongside types of A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is? theory, EduRev gives you an
ample number of questions to practice A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup