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A body executes simple harmonic motion under the action of force F1 with a time period 4/5s.If the force is changed to F2 it executes SHM with time period 3/5s .If both forces F1&F2 act simultaneously in the same direction on the body, it's time period will be a. 12/25s b. 24/25s c. 35/24s d. 15/12s?
Most Upvoted Answer
A body executes simple harmonic motion under the action of force F1 wi...
And F2 act simultaneously on the body, what will be the time period of the resulting SHM?
The time period of simple harmonic motion is given by:
T = 2π√(m/k)
where m is the mass of the body and k is the spring constant.
For the first case, T1 = 4/5s, so we have:
T1 = 2π√(m/k1)
Squaring both sides and rearranging, we get:
k1/m = (2π/T1)^2
For the second case, T2 = 3/5s, so we have:
T2 = 2π√(m/k2)
Squaring both sides and rearranging, we get:
k2/m = (2π/T2)^2
When both forces act simultaneously, the resulting force is:
F = F1 + F2
The spring constant of the resulting force is:
k = k1 + k2
The time period of the resulting SHM is:
T = 2π√(m/k)
Substituting for k and simplifying, we get:
T = 2π√(m/(k1 + k2))
= 2π√(m/((2π/T1)^2 + (2π/T2)^2))
= 2π√(m/(4π^2/T1^2 + 4π^2/T2^2))
= 2π√(m/4π^2(T1^2 + T2^2)/T1^2T2^2)
= 2π√(mT1^2T2^2/(T1^2 + T2^2))
Substituting the given values, we get:
T = 2π√(m(4/5)^2(3/5)^2/((4/5)^2 + (3/5)^2))
= 2π√(9m/41)
Therefore, the time period of the resulting SHM is 2π√(9m/41).
The time period of simple harmonic motion is given by:
T = 2π√(m/k)
where m is the mass of the body and k is the spring constant.
For the first case, T1 = 4/5s, so we have:
T1 = 2π√(m/k1)
Squaring both sides and rearranging, we get:
k1/m = (2π/T1)^2
For the second case, T2 = 3/5s, so we have:
T2 = 2π√(m/k2)
Squaring both sides and rearranging, we get:
k2/m = (2π/T2)^2
When both forces act simultaneously, the resulting force is:
F = F1 + F2
The spring constant of the resulting force is:
k = k1 + k2
The time period of the resulting SHM is:
T = 2π√(m/k)
Substituting for k and simplifying, we get:
T = 2π√(m/(k1 + k2))
= 2π√(m/((2π/T1)^2 + (2π/T2)^2))
= 2π√(m/(4π^2/T1^2 + 4π^2/T2^2))
= 2π√(m/4π^2(T1^2 + T2^2)/T1^2T2^2)
= 2π√(mT1^2T2^2/(T1^2 + T2^2))
Substituting the given values, we get:
T = 2π√(m(4/5)^2(3/5)^2/((4/5)^2 + (3/5)^2))
= 2π√(9m/41)
Therefore, the time period of the resulting SHM is 2π√(9m/41).
Community Answer
A body executes simple harmonic motion under the action of force F1 wi...
12/25..formula: T= T1 * T2/√(T1^2 + T2 ^2)
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A body executes simple harmonic motion under the action of force F1 with a time period 4/5s.If the force is changed to F2 it executes SHM with time period 3/5s .If both forces F1&F2 act simultaneously in the same direction on the body, it's time period will be a. 12/25s b. 24/25s c. 35/24s d. 15/12s?
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A body executes simple harmonic motion under the action of force F1 with a time period 4/5s.If the force is changed to F2 it executes SHM with time period 3/5s .If both forces F1&F2 act simultaneously in the same direction on the body, it's time period will be a. 12/25s b. 24/25s c. 35/24s d. 15/12s? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A body executes simple harmonic motion under the action of force F1 with a time period 4/5s.If the force is changed to F2 it executes SHM with time period 3/5s .If both forces F1&F2 act simultaneously in the same direction on the body, it's time period will be a. 12/25s b. 24/25s c. 35/24s d. 15/12s? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body executes simple harmonic motion under the action of force F1 with a time period 4/5s.If the force is changed to F2 it executes SHM with time period 3/5s .If both forces F1&F2 act simultaneously in the same direction on the body, it's time period will be a. 12/25s b. 24/25s c. 35/24s d. 15/12s?.
A body executes simple harmonic motion under the action of force F1 with a time period 4/5s.If the force is changed to F2 it executes SHM with time period 3/5s .If both forces F1&F2 act simultaneously in the same direction on the body, it's time period will be a. 12/25s b. 24/25s c. 35/24s d. 15/12s? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A body executes simple harmonic motion under the action of force F1 with a time period 4/5s.If the force is changed to F2 it executes SHM with time period 3/5s .If both forces F1&F2 act simultaneously in the same direction on the body, it's time period will be a. 12/25s b. 24/25s c. 35/24s d. 15/12s? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body executes simple harmonic motion under the action of force F1 with a time period 4/5s.If the force is changed to F2 it executes SHM with time period 3/5s .If both forces F1&F2 act simultaneously in the same direction on the body, it's time period will be a. 12/25s b. 24/25s c. 35/24s d. 15/12s?.
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A body executes simple harmonic motion under the action of force F1 with a time period 4/5s.If the force is changed to F2 it executes SHM with time period 3/5s .If both forces F1&F2 act simultaneously in the same direction on the body, it's time period will be a. 12/25s b. 24/25s c. 35/24s d. 15/12s?, a detailed solution for A body executes simple harmonic motion under the action of force F1 with a time period 4/5s.If the force is changed to F2 it executes SHM with time period 3/5s .If both forces F1&F2 act simultaneously in the same direction on the body, it's time period will be a. 12/25s b. 24/25s c. 35/24s d. 15/12s? has been provided alongside types of A body executes simple harmonic motion under the action of force F1 with a time period 4/5s.If the force is changed to F2 it executes SHM with time period 3/5s .If both forces F1&F2 act simultaneously in the same direction on the body, it's time period will be a. 12/25s b. 24/25s c. 35/24s d. 15/12s? theory, EduRev gives you an
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