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Answered on 14 Apr Learn Unit 10-Oscillation & Waves
Nazia Khanum
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:
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:
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:
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
Given that each wheel supports 750 kg and using the spring constant k=200000 N/mk=200000N/m, we can calculate the period TT. Then, we can solve for the damping constant bb.
Answered on 14 Apr Learn Unit 10-Oscillation & Waves
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'm here to shed light on a fundamental principle in classical mechanics. Let me demonstrate why, in the case of a particle undergoing linear Simple Harmonic Motion (SHM), the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
Firstly, let's delve into the nature of SHM. In linear SHM, a particle oscillates back and forth along a straight line, with its acceleration proportional and opposite to its displacement from a fixed equilibrium point. This leads to a sinusoidal motion characterized by a restoring force.
Now, to prove the equality of average kinetic and potential energies over a period, we must understand the expressions for kinetic and potential energies in SHM.
The kinetic energy (KE) of a particle is given by KE=12mv2KE=21mv2, where mm is the mass of the particle and vv is its velocity.
In SHM, velocity varies sinusoidally with displacement, reaching maximum at the equilibrium point and minimum at the extremities. Thus, the average kinetic energy over a period can be represented by 12mvmax221mvmax2, where vmaxvmax is the maximum velocity.
On the other hand, the potential energy (PE) of a particle undergoing SHM is given by PE=12kx2PE=21kx2, where kk is the spring constant and xx is the displacement from equilibrium.
In SHM, potential energy also varies sinusoidally, reaching maximum at the extremities and minimum at the equilibrium point. Hence, the average potential energy over a period can be represented by 12kxmax221kxmax2, where xmaxxmax is the maximum displacement.
Now, in SHM, the maximum displacement (xmaxxmax) and maximum velocity (vmaxvmax) occur at the same points in the motion, namely at the extremities. This is due to the relationship between displacement, velocity, and acceleration in SHM.
Since kinetic and potential energies both reach their maxima at the extremities, the average kinetic energy over a period equals the average potential energy over the same period.
Thus, in the realm of linear SHM, the balance between kinetic and potential energies is not only a theoretical construct but a practical reality, demonstrating the elegant harmony in the dynamics of oscillatory motion. If you're keen on further exploring such concepts or need assistance with related problems, don't hesitate to reach out. Remember, UrbanPro is your gateway to the best online coaching and tuition experiences!
Answered on 14 Apr Learn Unit 10-Oscillation & Waves
Nazia Khanum
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:
The restoring couple JJ can be related to the torque acting on the disc, which is given by:
J=I⋅αJ=I⋅α
Where:
The moment of inertia of a circular disc about its center is given by:
I=12mr2I=21mr2
Where:
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
Solving for αα:
α=(4.19)2×0.1125α=(4.19)2×0.1125 α≈1.86 Nm/radα≈1.86 Nm/rad
So, the torsional spring constant of the wire is approximately 1.86 Nm/rad1.86 Nm/rad.
Feel free to ask if you have any questions or need further clarification! And remember, UrbanPro is a great platform for finding excellent tutors for your academic needs.
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 14 Apr Learn Unit 10-Oscillation & Waves
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can help you tackle this problem step by step. Simple harmonic motion (SHM) is a fundamental concept in physics, and understanding it thoroughly can pave the way for mastering more complex topics.
Firstly, let's establish the key formulas for simple harmonic motion:
Displacement (x): x=A⋅sin(ωt+ϕ)x=A⋅sin(ωt+ϕ)
Velocity (v): v=A⋅ω⋅cos(ωt+ϕ)v=A⋅ω⋅cos(ωt+ϕ)
Acceleration (a): a=−A⋅ω2⋅sin(ωt+ϕ)a=−A⋅ω2⋅sin(ωt+ϕ)
Given that the amplitude AA is 5 cm and the period TT is 0.2 s, we can calculate the angular frequency (ωω) as 2π/T2π/T.
For x=5x=5 cm:
For x=3x=3 cm:
For x=0x=0 cm:
After finding the times for each displacement, we substitute them into the velocity and acceleration formulas to get the respective values.
UrbanPro provides a conducive environment for mastering such topics through personalized guidance and ample practice. Let's proceed step by step and delve into the intricacies of simple harmonic motion!
Answered on 14 Apr Learn Unit 10-Oscillation & Waves
Nazia Khanum
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.
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.
In SHM, the equation governing the motion of the mass is: x(t)=Acos(ωt+ϕ)x(t)=Acos(ωt+ϕ)
Where:
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.
The general equation for the velocity of an object undergoing SHM is: v(t)=−Aωsin(ωt+ϕ)v(t)=−Aωsin(ωt+ϕ)
At t=0t=0, the velocity of the mass is v0v0 towards the center, so: v(0)=−Aωsin(ϕ)=v0v(0)=−Aωsin(ϕ)=v0
At t=0t=0, the displacement of the mass is x0x0 from equilibrium, so: x(0)=Acos(ϕ)=x0x(0)=Acos(ϕ)=x0
We now have two equations: Acos(ϕ)=x0Acos(ϕ)=x0 −Aωsin(ϕ)=v0−Aωsin(ϕ)=v0
We can solve these equations simultaneously to find AA and ϕϕ: tan(ϕ)=−v0ωx0tan(ϕ)=−ωx0v0
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
x0
Thus, we've determined the amplitude AA in terms of the parameters ωω, x0x0, and v0v0.
In conclusion, using the principles of SHM and the given initial conditions, we've found the amplitude of the resulting oscillations in terms of ωω, x0x0, and v0v0. If you need further clarification or assistance, feel free to ask!
Answered on 14 Apr Learn Unit 10-Oscillation & Waves
Nazia Khanum
As a experienced tutor registered on UrbanPro, I can assure you that UrbanPro is indeed a great platform for finding online coaching and tuition services. Now, let's address your physics question.
Given:
We can calculate the speed of the transverse wave along the string using the formula:
v=Tμv=μT
Where:
The linear mass density μμ can be calculated as the mass per unit length of the string, given by:
μ=mLμ=Lm
Now, substituting the values:
μ=2.50 kg20.0 m=0.125 kg/mμ=20.0m2.50kg=0.125kg/m
v=200 N0.125 kg/mv=0.125kg/m200N
v=1600=40 m/sv=1600
=40m/s
Now, to find the time taken for the disturbance to reach the other end of the string, we can use the formula:
Time=DistanceSpeedTime=SpeedDistance
Since the disturbance travels the length of the string, the distance is L=20.0 mL=20.0m. So,
Time=20.0 m40 m/s=0.5 sTime=40m/s20.0m=0.5s
Therefore, the disturbance takes 0.5 s0.5s to reach the other end of the string. If you need further clarification or assistance, feel free to ask!
Take Class 11 Tuition from the Best Tutors
Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 14 Apr Learn Unit 10-Oscillation & Waves
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's tackle your physics problem.
We have a stone dropped from a tower, and we need to find out when the splash is heard at the top of the tower. We'll first determine the time it takes for the stone to fall to the pond and then the time it takes for the sound of the splash to travel back up to the top of the tower.
300=0×t+12×(−9.8)×t2300=0×t+21×(−9.8)×t2 300=−4.9t2300=−4.9t2 t2=3004.9t2=4.9300 t2≈61.22t2≈61.22 t≈61.22t≈61.22
t≈7.82t≈7.82 seconds (approximately)
So, it takes roughly 7.82 seconds for the stone to fall to the pond.
340=300t340=t300 t=300340t=340300 t≈0.88t≈0.88 seconds (approximately)
So, it takes roughly 0.88 seconds for the sound of the splash to travel up to the top of the tower.
Now, we add the time taken for the stone to fall to the pond and the time taken for the sound of the splash to travel up: Total time=7.82+0.88Total time=7.82+0.88 Total time≈8.7Total time≈8.7 seconds (approximately)
Therefore, the splash is heard at the top of the tower after approximately 8.7 seconds.
Answered on 14 Apr Learn Unit 10-Oscillation & Waves
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'd like to highlight that UrbanPro is indeed one of the best platforms for online coaching and tuition services. Now, let's address your physics question.
To find the tension in the steel wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20°C, we'll use the equation for the speed of a transverse wave on a stretched string:
v=Tμv=μT
Where:
First, let's find the linear mass density of the wire:
μ=mL=2.10 kg12.0 m=0.175 kg/mμ=Lm=12.0m2.10kg=0.175kg/m
Now, we'll use the given speed of sound in dry air at 20°C (vsound=340 m/svsound=340m/s) as the speed of the transverse wave (vv) and solve for the tension (TT):
340 m/s=T0.175 kg/m340m/s=0.175kg/mT
Squaring both sides to isolate TT, we get:
T=(340 m/s)2×0.175 kg/m=9520 NT=(340m/s)2×0.175kg/m=9520N
So, the tension in the wire should be 9520 N9520N for the speed of a transverse wave on the wire to equal the speed of sound in dry air at 20°C.
Answered on 14 Apr Learn Unit 10-Oscillation & Waves
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently assert that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into your question about travelling waves.
In the realm of physics, a travelling wave in one dimension is often represented by the function y=f(x,t)y=f(x,t), where xx and tt must appear in the combination x−vtx−vt or x+vtx+vt, i.e., y=f(x±vt)y=f(x±vt). This represents a wave moving either to the left or to the right with a velocity vv.
Now, to address your query regarding whether every function of x−vtx−vt or x+vtx+vt represents a travelling wave, we need to examine the characteristics of these functions.
A function of the form y=f(x−vt)y=f(x−vt) represents a wave moving to the right with velocity vv, while y=f(x+vt)y=f(x+vt) represents a wave moving to the left with velocity vv.
However, not every function of x−vtx−vt or x+vtx+vt necessarily represents a travelling wave. The key lies in the nature of the function f(x,t)f(x,t) itself. If the function f(x,t)f(x,t) is such that it describes a disturbance propagating through a medium with a constant velocity, then it represents a travelling wave.
To determine whether a given function for yy can possibly represent a travelling wave, we need to analyze its properties. If the function exhibits characteristics of a wave, such as periodicity, wavelength, and propagation, then it is likely to represent a travelling wave. Conversely, if the function lacks these characteristics or represents a stationary disturbance, then it may not correspond to a travelling wave.
In summary, while functions of x−vtx−vt or x+vtx+vt have the potential to represent travelling waves, it ultimately depends on the specific form and behavior of the function f(x,t)f(x,t). Further analysis and examination of the given functions for yy would be necessary to determine their nature as travelling waves.
Take Class 11 Tuition from the Best Tutors
Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 14 Apr Learn Unit 10-Oscillation & Waves
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'd be glad to help you with this physics problem!
When a sound wave encounters a boundary between two different mediums, such as air and water, it undergoes changes in its properties. Let's solve the problem step by step.
(a) The wavelength of the reflected sound: We can use the formula:
Given that the frequency of the sound emitted by the bat is 1000 kHz1000kHz and the speed of sound in air is 340 m/s340m/s, we can plug these values into the formula:
Now, calculate the wavelength in air.
(b) The wavelength of the transmitted sound: Similarly, we can use the same formula, but this time, we'll use the speed of sound in water, which is 1486 m/s1486m/s.
Now, calculate the wavelength in water.
Once you have both these values, you'll have the wavelengths of the reflected and transmitted sounds when the ultrasonic sound meets a water surface. If you need further clarification or assistance with the calculations, feel free to ask!
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