Concepts Of Physics MCQ Edition [Volume 1]
- Exam
- Concepts Of Physics MCQ Edition [Volume 1]
- Subject
- Physics
- Chapter
- Simple Harmonic Motion
Concepts Of Physics MCQ Edition [Volume 1] · Physics · Simple Harmonic Motion
Three springs A , B , and C , each having a force constant k , are...
Three springs , , and , each having a force constant , are attached to a particle of mass . Initially, the angle between any two springs is 120^{}^{{}^{\circ}}. If the particle is slightly pushed against spring and then released, determine its time period of oscillation.
Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
Let the displacement of the particle be towards spring .
The compression in spring is . The restoring force exerted by spring is , directed opposite to the displacement.
The angle between the line of action of spring and those of springs and is 120^{}^{{}^{\circ}}.
The extension in springs and is the component of displacement along their respective lengths, which is x \cos(180^{}^{{}^{\circ}} - 120^{}^{{}^{\circ}}) = x \cos 60^{}^{{}^{\circ}} = \dfrac{x}{2}.
The restoring force exerted by each of springs and is ABF_{A,B} = F_A \cos 60^{}^{{}^{\circ}} + F_B \cos 60^{}^{{}^{\circ}} = 2 \left(\dfrac{kx}{2}\right) \left(\dfrac{1}{2}\right) = \dfrac{kx}{2}
The total restoring force on the particle is:
\dfrac{kx}{2} = \dfrac{3kx}{2}
The effective spring constant of the system is \dfrac{3k}{2}.
The time period of oscillation is given by:
\sqrt{\dfrac{m}{k_{eff}}} ==2\pi \sqrt{\dfrac{2m}{3k}}
The compression in spring is . The restoring force exerted by spring is , directed opposite to the displacement.
The angle between the line of action of spring and those of springs and is 120^{}^{{}^{\circ}}.
The extension in springs and is the component of displacement along their respective lengths, which is x \cos(180^{}^{{}^{\circ}} - 120^{}^{{}^{\circ}}) = x \cos 60^{}^{{}^{\circ}} = \dfrac{x}{2}.
The restoring force exerted by each of springs and is ABF_{A,B} = F_A \cos 60^{}^{{}^{\circ}} + F_B \cos 60^{}^{{}^{\circ}} = 2 \left(\dfrac{kx}{2}\right) \left(\dfrac{1}{2}\right) = \dfrac{kx}{2}
The total restoring force on the particle is:
\dfrac{kx}{2} = \dfrac{3kx}{2}
The effective spring constant of the system is \dfrac{3k}{2}.
The time period of oscillation is given by:
\sqrt{\dfrac{m}{k_{eff}}} ==2\pi \sqrt{\dfrac{2m}{3k}}