Certainly! As a seasoned tutor on UrbanPro, I'd approach this problem step by step, ensuring a thorough understanding for my students.

Firstly, let's address the sagging of the suspension. When the entire automobile is placed on the suspension, it sags 15 cm. This sag can be attributed to the force exerted by the weight of the automobile, which is countered by the restoring force of the spring.

The force exerted by the weight of the automobile can be calculated using the formula:

Fweight=m×gFweight=m×g

Where:

• mm = mass of the automobile = 3000 kg
• gg = acceleration due to gravity = 10 m/s²

Fweight=3000 kg×10 m/s2Fweight=3000kg×10m/s2 Fweight=30000 NFweight=30000N

Now, to find the spring constant kk, we'll use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position:

Fspring=k×xFspring=k×x

Where:

• FspringFspring = force exerted by the spring
• kk = spring constant
• xx = displacement from equilibrium position

Given that the suspension sags 15 cm (or 0.15 m) under the weight of the automobile, we can set up the equation:

30000 N=k×0.15 m30000N=k×0.15m

From this, we can solve for kk:

k=30000 N0.15 mk=0.15m30000N k=200000 N/mk=200000N/m

So, the spring constant kk is 200000 N/m200000N/m.

Moving on to the damping constant bb, we're told that the amplitude of oscillation decreases by 50% during one complete oscillation. This indicates damping in the system.

The formula for damped oscillations involves the damping constant, and given that the amplitude decreases by 50% each oscillation, we can use this information to find bb.

The formula for the amplitude of damped oscillations is given by:

A=A0×e−b2mtA=A0×e−2mbt

Where:

• AA = amplitude at time tt
• A0A0 = initial amplitude
• bb = damping constant
• mm = mass of the system
• tt = time

Since the amplitude decreases by 50%, A=0.5A0A=0.5A0 after one complete oscillation. Substituting this into the formula:

0.5A0=A0×e−b2mT0.5A0=A0×e−2mbT

Where TT is the period of one complete oscillation.

0.5=e−b2mT0.5=e−2mbT

Taking the natural logarithm of both sides:

ln⁡(0.5)=−bT2mln(0.5)=−2mbT

We're given that the amplitude decreases by 50% during one complete oscillation. In simple harmonic motion, the period (TT) is related to the spring constant (kk) and the mass (mm) of the system:

T=2πmkT=2πkm

As an experienced tutor registered on UrbanPro, I'd be glad to assist you with this physics problem.

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

As a seasoned tutor on UrbanPro, I'd be delighted to guide you through this physics problem.

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

As an experienced tutor registered on UrbanPro, I'm delighted to help you with this question. UrbanPro indeed offers top-notch online coaching tuition services for students seeking academic assistance.

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