The period of oscillation of mass M, hanging from spring of force constant k is T. When additional mass m is attached to spring the period of oscillation becomes 5T/4. m/M=?

Explore Courses UPSC Exam UPSC Test Series Free Notes EduRev Infinity

Open App

Class 12 Exam > Class 12 Questions > The period of oscillation of mass M, hanging ...

Join the Discussion on the App

The period of oscillation of mass M, hanging from spring of force constant k is T. When additional mass m is attached to spring the period of oscillation becomes 5T/4. m/M=?

Join for Free

Join for Free

Most Upvoted Answer

The period of oscillation of mass M, hanging from spring of force cons...

Given:
- Period of oscillation with mass M: T
- Period of oscillation with mass M + m: 5T/4

Calculating the period of oscillation:
- The period of oscillation of a mass-spring system is given by the formula: T = 2π√(m/k), where m is the mass attached to the spring and k is the force constant of the spring.

Calculating period with mass M:
- Using the formula, for mass M: T = 2π√(M/k)

Calculating period with mass M + m:
- When m is added to mass M, the new period becomes 5T/4: 5T/4 = 2π√((M + m)/k)

Equating the two period equations:
- Equating the two period equations, we get: 2π√(M/k) = 2π√((M + m)/k)
- Simplifying the equation: √(M/k) = √((M + m)/k)
- Squaring both sides: M/k = (M + m)/k
- Solving for m: M = M + m
- Therefore, m/M = 1

Conclusion:
- The ratio of the additional mass m to the original mass M is 1.

Community Answer

The period of oscillation of mass M, hanging from spring of force cons...

T=2π√[m/k] use this formula...

Share via Whatsapp

View all answers Start learning for free

Explore Courses for Class 12 exam

Similar Class 12 Doubts

Read the following text and answer the following questions on the basis of the same: Roget’s spiral: Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor μ. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5–1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,...The force 10–3 N,is equivalent to

View answer

Read the following text and answer the following questions on the basis of the same:Roget’s spiral: Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor μ. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5–1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,...Magnetic effects

View answer

Read the following text and answer the following questions on the basis of the same: Roget’s spiral: Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor μ. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5–1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,...Why the spring shrinks in Roget’s spiral ?

View answer

Read the following text and answer the following questions on the basis of the same: Roget’s spiral: Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor μ. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5–1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,...What are the main 3 components in a Roget’s spiral?

View answer

Read the following text and answer the following questions on the basis of the same: Roget’s spiral: Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor μ. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5–1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,...What else can be used instead of mercury in Roget’s spiral ?

View answer

The period of oscillation of mass M, hanging from spring of force constant k is T. When additional mass m is attached to spring the period of oscillation becomes 5T/4. m/M=?

Share with a Friend

Answer this doubt

Question Description
The period of oscillation of mass M, hanging from spring of force constant k is T. When additional mass m is attached to spring the period of oscillation becomes 5T/4. m/M=? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The period of oscillation of mass M, hanging from spring of force constant k is T. When additional mass m is attached to spring the period of oscillation becomes 5T/4. m/M=? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The period of oscillation of mass M, hanging from spring of force constant k is T. When additional mass m is attached to spring the period of oscillation becomes 5T/4. m/M=?.

Solutions for The period of oscillation of mass M, hanging from spring of force constant k is T. When additional mass m is attached to spring the period of oscillation becomes 5T/4. m/M=? in English & in Hindi are available as part of our courses for Class 12. Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.

Here you can find the meaning of The period of oscillation of mass M, hanging from spring of force constant k is T. When additional mass m is attached to spring the period of oscillation becomes 5T/4. m/M=? defined & explained in the simplest way possible. Besides giving the explanation of The period of oscillation of mass M, hanging from spring of force constant k is T. When additional mass m is attached to spring the period of oscillation becomes 5T/4. m/M=?, a detailed solution for The period of oscillation of mass M, hanging from spring of force constant k is T. When additional mass m is attached to spring the period of oscillation becomes 5T/4. m/M=? has been provided alongside types of The period of oscillation of mass M, hanging from spring of force constant k is T. When additional mass m is attached to spring the period of oscillation becomes 5T/4. m/M=? theory, EduRev gives you an ample number of questions to practice The period of oscillation of mass M, hanging from spring of force constant k is T. When additional mass m is attached to spring the period of oscillation becomes 5T/4. m/M=? tests, examples and also practice Class 12 tests.

Explore Courses for Class 12 exam