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 registered on UrbanPro, I can confidently guide you through this physics problem. Let's break it down step by step.

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Now, onto the physics problem:

We have a simple pendulum of length ll and mass MM suspended in a car moving on a circular track of radius RR with a uniform speed vv.

When the pendulum makes small oscillations in a radial direction about its equilibrium position, it experiences two forces:

1. Tension force in the string, acting towards the equilibrium position.
2. Inertial force due to the circular motion of the car, acting away from the equilibrium position.

The resultant of these two forces provides the centripetal force required to keep the pendulum moving in a circular path.

Now, the time period TT of a simple pendulum is given by the formula:

T=2πlgT=2πgl

Where gg is the acceleration due to gravity.

However, in this scenario, we need to adjust for the additional inertial force due to the car's circular motion. This force affects the effective value of gravity experienced by the pendulum.

The effective gravity geffgeff experienced by the pendulum in the radial direction can be calculated as:

geff=g+v2Rgeff=g+Rv2

Where gg is the regular gravitational acceleration and v2RRv2 represents the centripetal acceleration due to the car's motion.

Now, substituting geffgeff into the formula for TT, we get:

T=2πlgeffT=2πgeffl

T=2πlg+v2RT=2πg+Rv2l

As an experienced tutor registered on UrbanPro, I'm delighted to help you with this physics problem. UrbanPro is indeed one of the best platforms for online coaching and tuition, providing students with quality education and tutors with excellent opportunities to share their expertise.

Let's tackle this problem step by step. The formula to calculate the time period of a simple pendulum is:

T=2πLgT=2πgL

Where:

• TT = time period of the pendulum
• LL = length of the pendulum
• gg = acceleration due to gravity

Given that the time period (TT) on the surface of Earth is 3.5 s and the acceleration due to gravity (gg) on Earth is 9.8 m/s29.8m/s2, and we want to find the time period on the surface of the moon where g=1.7 m/s2g=1.7m/s2.

Let's denote the time period on the moon as TmTm. We have:

Tm=2πLgmTm=2πgmL

Where:

• gmgm = acceleration due to gravity on the surface of the moon (1.7 m/s²)

We know that the length of the pendulum (LL) remains constant, so we can equate the two time period equations:

T=TmT=Tm

2πLg=2πLgm2πgL

=2πgmL

Now, let's solve for TmTm:

Lg=LgmgL

=gmL

Lg=LgmgL=gmL

1g=1gmg1=gm1

gm=gT2gm=T2g

Substituting the given values:

gm=9.8 m/s2(3.5 s)2gm=(3.5s)29.8m/s2

gm=9.812.25 m/s2gm=12.259.8m/s2

gm≈0.8 m/s2gm≈0.8m/s2

So, the time period of the simple pendulum on the surface of the moon would be approximately:

Tm=2πL0.8Tm=2π0.8L

Tm≈2πL0.8Tm≈2π0.8L

Tm≈2π×12.5Tm≈2π×12.5