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Given that a circular disc of mass 10 kg is suspended by a wire attached to its center, and the wire is twisted by rotating the disc and released, resulting in torsional oscillations with a period of 1.5 seconds. The radius of the disc is 15 cm.

To determine the torsional spring constant (α) of the wire, we can use the equation provided:

J=−αθJ=−αθ

Where:

• JJ is the restoring couple,
• θθ is the angle of twist.

The restoring couple JJ can be related to the torque acting on the disc, which is given by:

J=I⋅αJ=I⋅α

Where:

• II is the moment of inertia of the disc about its center.

The moment of inertia of a circular disc about its center is given by:

I=12mr2I=21mr2

Where:

• mm is the mass of the disc,
• rr is the radius of the disc.

Given that m=10m=10 kg and r=15r=15 cm, we can calculate the moment of inertia II.

I=12×10×(0.15)2I=21×10×(0.15)2 I=0.1125 kg m2I=0.1125 kg m2

Now, substituting II into the equation for the restoring couple:

J=0.1125×αJ=0.1125×α

Since the period of torsional oscillations (TT) is related to the angular frequency (ωω) by T=2πωT=ω2π, we can find ωω:

T=2πωT=ω2π ω=2πTω=T2π ω=2π1.5ω=1.52π ω≈4.19 rad/sω≈4.19 rad/s

Now, the relation between angular frequency (ωω) and torsional spring constant (αα) is:

ω=αIω=Iα

Substituting the known values:

4.19=α0.11254.19=0.1125α

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When a mass attached to a spring is free to oscillate without friction or damping, we can model its motion using simple harmonic motion (SHM) principles. Let's break down the problem step by step.

1. At time t=0t=0, the mass is pulled to a distance x0x0 from its equilibrium position and given an initial velocity v0v0 towards the center.

2. In SHM, the equation governing the motion of the mass is: x(t)=Acos⁡(ωt+ϕ)x(t)=Acos(ωt+ϕ)

Where:

• x(t)x(t) is the displacement from equilibrium at time tt,
• AA is the amplitude (the maximum displacement from equilibrium),
• ωω is the angular frequency (related to the angular velocity),
• ϕϕ is the phase angle.

3. Since the mass is initially displaced from equilibrium and given an initial velocity, we'll need to determine the amplitude AA in terms of ωω, x0x0, and v0v0.

4. The general equation for the velocity of an object undergoing SHM is: v(t)=−Aωsin⁡(ωt+ϕ)v(t)=−Aωsin(ωt+ϕ)

5. At t=0t=0, the velocity of the mass is v0v0 towards the center, so: v(0)=−Aωsin⁡(ϕ)=v0v(0)=−Aωsin(ϕ)=v0

6. At t=0t=0, the displacement of the mass is x0x0 from equilibrium, so: x(0)=Acos⁡(ϕ)=x0x(0)=Acos(ϕ)=x0

7. We now have two equations: Acos⁡(ϕ)=x0Acos(ϕ)=x0 −Aωsin⁡(ϕ)=v0−Aωsin(ϕ)=v0

8. We can solve these equations simultaneously to find AA and ϕϕ: tan⁡(ϕ)=−v0ωx0tan(ϕ)=−ωx0v0

9. Once we find ϕϕ, we can substitute it back into one of the equations to find AA: A=x0cos⁡(ϕ)=x01+(v0ωx0)2A=cos(ϕ)x0=1+(ωx0v0)2

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Let's delve into the physics problem at hand. We have an air chamber with a volume VV and a neck area of cross-section into which a ball of mass mm just fits and can move up and down without any friction. When the ball is pressed down a little and released, it executes Simple Harmonic Motion (SHM). We need to derive an expression for the time period of oscillations, assuming pressure-volume variations of air to be isothermal.

To begin, let's analyze the forces acting on the ball. When the ball is pushed down, it experiences an upward force due to the buoyant force and a downward force due to gravity. At equilibrium, these forces balance out, and the ball remains stationary. When the ball is displaced slightly downwards and released, it experiences an upward force due to the compressed air in the chamber, leading it to oscillate.

The restoring force acting on the ball is due to the pressure difference between the compressed air below the ball and the less compressed air above it. According to Boyle's Law for isothermal processes, PV=constantPV=constant, where PP is pressure and VV is volume. Thus, when the ball is displaced downwards by a distance xx, the volume of air below it decreases, causing an increase in pressure, which provides the restoring force.

Using Hooke's Law, which states that the restoring force is directly proportional to the displacement, we can write:

F=−kxF=−kx

Where kk is the spring constant.

The pressure difference ΔPΔP across the ball can be expressed as:

ΔP=P0−P1ΔP=P0−P1

Where P0P0 is the pressure when the ball is at its equilibrium position, and P1P1 is the pressure when the ball is displaced by xx.

We can express P0P0 and P1P1 using Boyle's Law:

P0=kVandP1=kV−xP0=VkandP1=V−xk

Thus, the pressure difference ΔPΔP is:

ΔP=kV−kV−xΔP=Vk−V−xk

The force exerted by this pressure difference on the ball is:

F=AΔPF=AΔP

Where AA is the cross-sectional area of the neck.

So, we have:

F=A(kV−kV−x)F=A(Vk−V−xk)

F=kA(1V−1V−x)F=kA(V1−V−x1)

Now, equating this to −kx−kx (according to Hooke's Law), we get:

kA(1V−1V−x)=−kxkA(V1−V−x1)=−kx

A(1V−1V−x)=−xA(V1−V−x1)=−x

AV−AxV(V−x)=−xVA−V(V−x)Ax=−x

AV−Ax=−x(V−x)AV−Ax=−x(V−x)

AV−Ax=−xV+x2AV−Ax=−xV+x2

AV=−xV+Ax+x2AV=−xV+Ax+x2

x2−Ax+AV=0x2−Ax+AV=0

This is a quadratic equation in xx. Solving this equation will give us the value of xx, which is the amplitude of the oscillation. Then, we can use the formula for the time period of SHM:

T=2πmkT=2πkm