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A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period T. If the mass is increased by m, the time period becomes 5T/3. then the ratio of m/M is
- a)9/ 16
- b)25/ 16
- c)3/9
- d)25/22
Correct answer is option 'D'. Can you explain this answer?
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A mass M is suspended from a spring of negligible mass. The spring is ...
Given:
- Mass of the spring: M
- Additional mass added: m
- Time period before adding the mass: T
- Time period after adding the mass: 5T/3
To find:
The ratio of m/M
Explanation:
1. Time Period of Simple Harmonic Motion:
The time period of simple harmonic motion is given by the formula:
T = 2π√(m/k)
where T is the time period, m is the mass, and k is the spring constant.
2. Effect of Mass on Time Period:
From the equation for the time period, we can see that the time period is inversely proportional to the square root of the mass. This means that as the mass increases, the time period increases.
3. Determining the New Time Period:
Let's assume the spring constant of the spring is k₁ before adding the mass and k₂ after adding the mass. Since the spring is pulled a little and then released, we can assume that the extension of the spring is directly proportional to the added mass.
Therefore, we can write:
m/k₁ = m/(k₁ + k₂)
Simplifying the equation:
k₁ = k₁ + k₂
This implies that k₂ = 0, which means the spring constant after adding the mass is negligible.
4. Relationship between Time Period and Mass:
Using the equation for the time period:
T = 2π√(m/k)
We can rewrite it as:
T ∝ √(1/m)
This indicates that the time period is inversely proportional to the square root of the mass.
5. Determining the Ratio of m/M:
Since the time period after adding the mass is 5T/3, we can write:
5T/3 = 2π√((M+m)/k)
Simplifying the equation:
(M+m)/k = (5T/3)^2
Using the relationship between time period and mass:
(M+m)/M = (√(M+m)/√M)^2 = (5T/3T)^2 = (5/3)^2
Simplifying further:
(M+m)/M = 25/9
Therefore, the ratio of m/M is 16/9.
Answer:
The ratio of m/M is 16/9. (Option A)
- Mass of the spring: M
- Additional mass added: m
- Time period before adding the mass: T
- Time period after adding the mass: 5T/3
To find:
The ratio of m/M
Explanation:
1. Time Period of Simple Harmonic Motion:
The time period of simple harmonic motion is given by the formula:
T = 2π√(m/k)
where T is the time period, m is the mass, and k is the spring constant.
2. Effect of Mass on Time Period:
From the equation for the time period, we can see that the time period is inversely proportional to the square root of the mass. This means that as the mass increases, the time period increases.
3. Determining the New Time Period:
Let's assume the spring constant of the spring is k₁ before adding the mass and k₂ after adding the mass. Since the spring is pulled a little and then released, we can assume that the extension of the spring is directly proportional to the added mass.
Therefore, we can write:
m/k₁ = m/(k₁ + k₂)
Simplifying the equation:
k₁ = k₁ + k₂
This implies that k₂ = 0, which means the spring constant after adding the mass is negligible.
4. Relationship between Time Period and Mass:
Using the equation for the time period:
T = 2π√(m/k)
We can rewrite it as:
T ∝ √(1/m)
This indicates that the time period is inversely proportional to the square root of the mass.
5. Determining the Ratio of m/M:
Since the time period after adding the mass is 5T/3, we can write:
5T/3 = 2π√((M+m)/k)
Simplifying the equation:
(M+m)/k = (5T/3)^2
Using the relationship between time period and mass:
(M+m)/M = (√(M+m)/√M)^2 = (5T/3T)^2 = (5/3)^2
Simplifying further:
(M+m)/M = 25/9
Therefore, the ratio of m/M is 16/9.
Answer:
The ratio of m/M is 16/9. (Option A)
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A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period T. If the mass is increased by m, the time period becomes 5T/3. then the ratio of m/M isa)9/ 16b)25/ 16c)3/9d)25/22Correct answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period T. If the mass is increased by m, the time period becomes 5T/3. then the ratio of m/M isa)9/ 16b)25/ 16c)3/9d)25/22Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period T. If the mass is increased by m, the time period becomes 5T/3. then the ratio of m/M isa)9/ 16b)25/ 16c)3/9d)25/22Correct answer is option 'D'. Can you explain this answer?.
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