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An ant sitting on a string can pull the string with a maximum force up to three times of its own weight. A = 10/(pi ^ 2) * m A sine wave of amplitude in vertical plane is generated in the string. Find the minimum time period of oscillations of string in second so that ant will not fall from the string. (Neglect the variation of tension in the string due to weight of the ant, g=10m/s^ 2 )?
Most Upvoted Answer
An ant sitting on a string can pull the string with a maximum force up...
F=ma
F=3mg
a=-w2A
therefore 3mg=mw2A
w=root3xpi
t=2/root3
t=1.176
F=3mg
a=-w2A
therefore 3mg=mw2A
w=root3xpi
t=2/root3
t=1.176
Community Answer
An ant sitting on a string can pull the string with a maximum force up...
Problem Statement:
An ant is sitting on a string and can pull the string with a maximum force up to three times its own weight. A sine wave of amplitude in the vertical plane is generated in the string. We need to find the minimum time period of the oscillations of the string so that the ant will not fall from the string. We can neglect the variation of tension in the string due to the weight of the ant and consider the acceleration due to gravity as 10 m/s^2.
Solution:
To solve this problem, we need to consider the forces acting on the ant when the string is oscillating. The ant can exert a maximum force of three times its own weight, which means the maximum force it can exert is 3mg, where m is the mass of the ant and g is the acceleration due to gravity.
Forces acting on the ant:
- Tension in the string: The tension in the string is responsible for providing the centripetal force required to keep the ant moving in a circular path. Let's assume the tension in the string is T.
- Weight of the ant: The weight of the ant is acting vertically downwards and can be calculated as mg.
Conditions for the ant not to fall:
For the ant not to fall from the string, the tension in the string should be greater than or equal to the weight of the ant. Mathematically, we can write this as T ≥ mg.
Centripetal force:
The tension in the string also provides the centripetal force required to keep the ant in circular motion. The centripetal force can be calculated as T = (m * v^2) / r, where v is the velocity of the ant and r is the radius of the circular path.
Relation between velocity and time period:
The velocity of the ant can be related to the time period of the oscillation using the equation v = 2πA / T, where A is the amplitude of the oscillation and T is the time period.
Minimum time period:
To find the minimum time period, we need to find the maximum value of T for which the ant does not fall. From the condition T ≥ mg, we can substitute T = 3mg (maximum force the ant can exert) and solve for T.
Substituting the values:
T = 3mg
T = (m * v^2) / r
3mg = (m * (2πA / T)^2) / r
Simplifying the equation, we get:
3gr = 4π^2A^2
Minimum time period:
The time period can be calculated using the equation T = 2π√(r / g), where r is the radius of the circular path and g is the acceleration due to gravity. We can substitute this value of T in the above equation and solve for r.
3g(2π√(r / g)) = 4π^2A^2
6√(rg) = 4πA^2
√(rg) = 2/3πA^2
rg = 4/9π^2A^4
r = 4/9π^2A^4g
Substituting
An ant is sitting on a string and can pull the string with a maximum force up to three times its own weight. A sine wave of amplitude in the vertical plane is generated in the string. We need to find the minimum time period of the oscillations of the string so that the ant will not fall from the string. We can neglect the variation of tension in the string due to the weight of the ant and consider the acceleration due to gravity as 10 m/s^2.
Solution:
To solve this problem, we need to consider the forces acting on the ant when the string is oscillating. The ant can exert a maximum force of three times its own weight, which means the maximum force it can exert is 3mg, where m is the mass of the ant and g is the acceleration due to gravity.
Forces acting on the ant:
- Tension in the string: The tension in the string is responsible for providing the centripetal force required to keep the ant moving in a circular path. Let's assume the tension in the string is T.
- Weight of the ant: The weight of the ant is acting vertically downwards and can be calculated as mg.
Conditions for the ant not to fall:
For the ant not to fall from the string, the tension in the string should be greater than or equal to the weight of the ant. Mathematically, we can write this as T ≥ mg.
Centripetal force:
The tension in the string also provides the centripetal force required to keep the ant in circular motion. The centripetal force can be calculated as T = (m * v^2) / r, where v is the velocity of the ant and r is the radius of the circular path.
Relation between velocity and time period:
The velocity of the ant can be related to the time period of the oscillation using the equation v = 2πA / T, where A is the amplitude of the oscillation and T is the time period.
Minimum time period:
To find the minimum time period, we need to find the maximum value of T for which the ant does not fall. From the condition T ≥ mg, we can substitute T = 3mg (maximum force the ant can exert) and solve for T.
Substituting the values:
T = 3mg
T = (m * v^2) / r
3mg = (m * (2πA / T)^2) / r
Simplifying the equation, we get:
3gr = 4π^2A^2
Minimum time period:
The time period can be calculated using the equation T = 2π√(r / g), where r is the radius of the circular path and g is the acceleration due to gravity. We can substitute this value of T in the above equation and solve for r.
3g(2π√(r / g)) = 4π^2A^2
6√(rg) = 4πA^2
√(rg) = 2/3πA^2
rg = 4/9π^2A^4
r = 4/9π^2A^4g
Substituting
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An ant sitting on a string can pull the string with a maximum force up to three times of its own weight. A = 10/(pi ^ 2) * m A sine wave of amplitude in vertical plane is generated in the string. Find the minimum time period of oscillations of string in second so that ant will not fall from the string. (Neglect the variation of tension in the string due to weight of the ant, g=10m/s^ 2 )?
Question Description
An ant sitting on a string can pull the string with a maximum force up to three times of its own weight. A = 10/(pi ^ 2) * m A sine wave of amplitude in vertical plane is generated in the string. Find the minimum time period of oscillations of string in second so that ant will not fall from the string. (Neglect the variation of tension in the string due to weight of the ant, g=10m/s^ 2 )? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An ant sitting on a string can pull the string with a maximum force up to three times of its own weight. A = 10/(pi ^ 2) * m A sine wave of amplitude in vertical plane is generated in the string. Find the minimum time period of oscillations of string in second so that ant will not fall from the string. (Neglect the variation of tension in the string due to weight of the ant, g=10m/s^ 2 )? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ant sitting on a string can pull the string with a maximum force up to three times of its own weight. A = 10/(pi ^ 2) * m A sine wave of amplitude in vertical plane is generated in the string. Find the minimum time period of oscillations of string in second so that ant will not fall from the string. (Neglect the variation of tension in the string due to weight of the ant, g=10m/s^ 2 )?.
An ant sitting on a string can pull the string with a maximum force up to three times of its own weight. A = 10/(pi ^ 2) * m A sine wave of amplitude in vertical plane is generated in the string. Find the minimum time period of oscillations of string in second so that ant will not fall from the string. (Neglect the variation of tension in the string due to weight of the ant, g=10m/s^ 2 )? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An ant sitting on a string can pull the string with a maximum force up to three times of its own weight. A = 10/(pi ^ 2) * m A sine wave of amplitude in vertical plane is generated in the string. Find the minimum time period of oscillations of string in second so that ant will not fall from the string. (Neglect the variation of tension in the string due to weight of the ant, g=10m/s^ 2 )? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ant sitting on a string can pull the string with a maximum force up to three times of its own weight. A = 10/(pi ^ 2) * m A sine wave of amplitude in vertical plane is generated in the string. Find the minimum time period of oscillations of string in second so that ant will not fall from the string. (Neglect the variation of tension in the string due to weight of the ant, g=10m/s^ 2 )?.
Solutions for An ant sitting on a string can pull the string with a maximum force up to three times of its own weight. A = 10/(pi ^ 2) * m A sine wave of amplitude in vertical plane is generated in the string. Find the minimum time period of oscillations of string in second so that ant will not fall from the string. (Neglect the variation of tension in the string due to weight of the ant, g=10m/s^ 2 )? in English & in Hindi are available as part of our courses for JEE.
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