JEE Exam > JEE Questions > A point mass m=20kg, is suspended by a massle...
Start Learning for Free
A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these?
Most Upvoted Answer
A point mass m=20kg, is suspended by a massless spring of constant 200...
Analysis:
The motion of the point mass suspended by the spring can be modeled using simple harmonic motion principles. The equation of motion for a mass-spring system is given by y(t) = A * sin(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase angle.
Given:
- Mass (m) = 20 kg
- Spring constant (k) = 2000 N/m
- Elongation in the spring = 15 cm = 0.15 m
- Acceleration due to gravity (g) = 10 m/s^2
Calculations:
1. Calculate the angular frequency (ω) using the formula: ω = √(k/m)
ω = √(2000/20) = 10 rad/s
2. Determine the amplitude (A) using the elongation in the spring: A = 0.15 m
3. Since the point mass is released from rest, the initial phase angle (φ) is 0.
Equation of motion:
Therefore, the equation of displacement of the particle as a function of time is:
y(t) = 0.15 * sin(10t)
Conclusion:
Thus, the correct equation representing the displacement of the point mass as a function of time is y(t) = 0.15 * sin(10t), which corresponds to option A.
The motion of the point mass suspended by the spring can be modeled using simple harmonic motion principles. The equation of motion for a mass-spring system is given by y(t) = A * sin(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase angle.
Given:
- Mass (m) = 20 kg
- Spring constant (k) = 2000 N/m
- Elongation in the spring = 15 cm = 0.15 m
- Acceleration due to gravity (g) = 10 m/s^2
Calculations:
1. Calculate the angular frequency (ω) using the formula: ω = √(k/m)
ω = √(2000/20) = 10 rad/s
2. Determine the amplitude (A) using the elongation in the spring: A = 0.15 m
3. Since the point mass is released from rest, the initial phase angle (φ) is 0.
Equation of motion:
Therefore, the equation of displacement of the particle as a function of time is:
y(t) = 0.15 * sin(10t)
Conclusion:
Thus, the correct equation representing the displacement of the point mass as a function of time is y(t) = 0.15 * sin(10t), which corresponds to option A.
Community Answer
A point mass m=20kg, is suspended by a massless spring of constant 200...
(C)
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
|
|
Top Courses for JEEView all
A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these?
Question Description
A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these?.
A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these?.
Solutions for A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these? 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 point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these? defined & explained in the simplest way possible. Besides giving the explanation of
A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these?, a detailed solution for A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these? has been provided alongside types of A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these? theory, EduRev gives you an
ample number of questions to practice A point mass m=20kg, is suspended by a massless spring of constant 2000 N/m.The point mass is released when elongation in the spring is 15cm. The equation of displacement of partical as a function of time is (take g=10m/s^2) A. y=10sin10t. B . y=10cos10t C. y=10sin(10t π/6) D. None of these? 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
|
Follow Us
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup