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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.
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!
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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'd be glad to explain this concept to you. Let's delve into the physics behind this scenario.
When a U-tube containing mercury is connected to a suction pump and the other end to the atmosphere, a small pressure difference is maintained between the two columns. This pressure difference causes the mercury levels in the two arms of the U-tube to be different, with one side being higher than the other.
Now, when the suction pump is removed, the pressure inside the U-tube equalizes with the atmospheric pressure. As a result, the mercury in the higher arm of the U-tube begins to fall while the mercury in the lower arm rises until both levels stabilize at the same height.
However, due to the inertia of the mercury, it doesn't stop immediately but overshoots the equilibrium position, creating a restoring force that brings it back towards the equilibrium position. This process repeats, causing the column of mercury to oscillate back and forth around the equilibrium position.
This oscillatory motion of the column of mercury in the U-tube is essentially simple harmonic motion (SHM). SHM occurs when a restoring force is proportional to the displacement from the equilibrium position and acts in the opposite direction to the displacement.
In the case of the U-tube, the restoring force is provided by gravity pulling the mercury back towards the equilibrium position. The displacement from the equilibrium position is directly proportional to the pressure difference between the two arms of the U-tube. Thus, the motion of the column of mercury in the U-tube can be described as simple harmonic motion.
Understanding this phenomenon not only helps in grasping the concept of SHM but also provides insights into fluid dynamics and pressure systems. If you have any further questions or need clarification on any point, feel free to ask!
Answered on 14 Apr Learn Unit 10-Oscillation & Waves
Nazia Khanum
Ah, buoyancy and oscillations, always fascinating topics! Here's how we can tackle this problem step by step, showcasing why UrbanPro is your go-to for online coaching tuition.
Firstly, let's establish some key concepts:
Buoyant Force: When the cork is submerged in the liquid, it experiences an upward buoyant force equal to the weight of the liquid displaced by the cork, given by the formula: Fbuoyant=ρ1gVFbuoyant=ρ1gV, where ρ1ρ1 is the density of the liquid, gg is the acceleration due to gravity, and VV is the volume of the submerged part of the cork.
Weight of the Cork: The weight of the cork can be calculated as Fweight=mgFweight=mg, where mm is the mass of the cork and gg is the acceleration due to gravity.
Now, when the cork is slightly depressed and released, it oscillates up and down due to the restoring force provided by the buoyant force and the weight of the cork.
The net force acting on the cork is the difference between the buoyant force and the weight of the cork:
Fnet=Fbuoyant−FweightFnet=Fbuoyant−Fweight
Substituting the expressions for FbuoyantFbuoyant and FweightFweight, we get:
Fnet=ρ1gV−mgFnet=ρ1gV−mg
Now, to find the volume of the submerged part of the cork (VV), we use the formula for the volume of a cylinder:
V=Ah′V=Ah′
Where AA is the base area of the cork and h′h′ is the depth to which the cork is submerged.
The depth h′h′ can be expressed as h−xh−x, where hh is the total height of the cork and xx is the depression from its equilibrium position.
Now, let's substitute VV into our equation for FnetFnet:
Fnet=ρ1gA(h−x)−mgFnet=ρ1gA(h−x)−mg
This equation represents the net force acting on the cork as a function of its displacement xx. Since it's a linear spring-like force, the motion will be simple harmonic.
We can apply Newton's second law to this system to derive the equation of motion and then find the period of oscillation.
Fnet=maFnet=ma
Where aa is the acceleration of the cork. Substituting FnetFnet into this equation and rearranging, we get:
ρ1gA(h−x)−mg=md2xdt2ρ1gA(h−x)−mg=mdt2d2x
From here, we can solve for xx to find the equation of motion, and subsequently, find the period of oscillation using the standard formula for simple harmonic motion.
And voilà! UrbanPro provides the best online coaching tuition to help you understand and master such intricate concepts with ease. Feel free to reach out for further clarification or assistance!
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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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 14 Apr Learn Unit 10-Oscillation & Waves
Nazia Khanum
Sure! Let's tackle this problem step by step.
(a) To find the speed of a transverse wave on the string, we can use the formula:
v=Tμv=μT
Where:
Given that the frequency of the fundamental mode is f=45f=45 Hz, we know that f=1Tf=T1, where TT is the period of the wave. Since the wire is vibrating in its fundamental mode, the frequency is the same as the frequency of the wave. So, T=1fT=f1.
Let's first find the period of the wave: T=145T=451 seconds.
Now, let's use the formula for wave speed:
v=Tμv=μT
v=1454.0×10−2v=4.0×10−2451
v=145×4.0×10−2v=45×4.0×10−21
v=11.8v=1.81
v=0.55556v=0.55556
v≈0.745 m/sv≈0.745m/s
So, the speed of the transverse wave on the string is approximately 0.745 m/s0.745m/s.
(b) Now, let's find the tension in the string. We'll use the formula:
T=μv2T=μv2
T=(4.0×10−2)×(0.745)2T=(4.0×10−2)×(0.745)2
T=4.0×10−2×0.55556T=4.0×10−2×0.55556
T≈0.0222 NT≈0.0222N
So, the tension in the string is approximately 0.0222 N0.0222N.
If you have any further questions or need clarification, feel free to ask! And remember, UrbanPro is here to support your learning journey.
Answered on 14 Apr Learn Unit 10-Oscillation & Waves
Nazia Khanum
Certainly! Let's break down the problem step by step.
Firstly, we need to understand the concept of resonance. In this scenario, the tube resonates with a fixed frequency source, which is a tuning fork with a frequency of 340 Hz. The resonance occurs when the length of the tube is such that it allows for a standing wave to form, with nodes at both ends and an antinode in the middle.
Given that the tube length for resonance is observed at 25.5 cm and 79.3 cm, we can use the formula for the fundamental frequency of a closed-open pipe to find the speed of sound.
The formula for the fundamental frequency of a closed-open pipe is:
f=v2Lf=2Lv
Where:
We have two sets of data for LL, which gives us two equations:
f1=v2L1f1=2L1v f2=v2L2f2=2L2v
We can rearrange these equations to solve for vv:
v=2f1L1v=2f1L1 v=2f2L2v=2f2L2
Now, we can average the values of vv obtained from these two equations to get a more accurate estimation of the speed of sound.
vavg=(2f1L1+2f2L2)2vavg=2(2f1L1+2f2L2)
Plugging in the values:
vavg=(2×340 Hz×0.255 m+2×340 Hz×0.793 m)2vavg=2(2×340Hz×0.255m+2×340Hz×0.793m)
vavg=(340×0.255+340×0.793)2vavg=2(340×0.255+340×0.793)
vavg=(86.7+269.62)2vavg=2(86.7+269.62)
vavg=356.322vavg=2356.32
vavg=178.16 m/svavg=178.16m/s
So, the estimated speed of sound in air at the temperature of the experiment is approximately 178.16 m/s178.16m/s.
In real-life scenarios, this value might deviate slightly due to factors like temperature and humidity affecting the speed of sound, but this calculation provides a good approximation based on the given data.
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 question.
When a steel rod is clamped at its midpoint, it effectively acts as a closed-closed pipe, where the fundamental frequency (ff) of longitudinal vibrations can be given by the equation:
f=v2Lf=2Lv
Where:
Rearranging the equation to solve for vv:
v=2Lfv=2Lf
Substituting the given values:
v=2×1 m×2530 Hzv=2×1m×2530Hz
v=5060 m/sv=5060m/s
So, the speed of sound in steel is 5060 meters per second. If you need further explanation or assistance, feel free to ask!
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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 assure you that UrbanPro is one of the best platforms for finding online coaching and tuition services. Now, let's address your question.
For a closed pipe, such as the one described here, the fundamental frequency (first harmonic) is given by:
f1=v2Lf1=2Lv
Where:
Plugging in the values:
f1=3402×0.2=850 Hzf1=2×0.2340=850 Hz
Now, let's find out which harmonic mode of the pipe is resonantly excited by a 430 Hz source. For a closed pipe, the resonant frequencies are odd harmonics. So, we can find the nearest odd multiple of the fundamental frequency to 430 Hz:
fn=(2n−1)f1fn=(2n−1)f1
430 Hz≈(2n−1)×850 Hz430 Hz≈(2n−1)×850 Hz
n≈430850+12n≈850430+21
n≈0.505n≈0.505
Since nn must be an integer for the harmonic mode, we take the nearest integer, which is n=1n=1. So, the first harmonic mode (fundamental frequency) of the pipe is resonantly excited by a 430 Hz source.
Now, let's consider if both ends of the pipe are open. For an open pipe, the fundamental frequency (first harmonic) is given by:
f1=v2Lf1=2Lv
Plugging in the values:
f1=3402×0.2=850 Hzf1=2×0.2340=850 Hz
Now, for an open pipe, the resonant frequencies are all harmonics (both odd and even). So, the same 430 Hz source will not resonate with the pipe because it does not match any of the resonant frequencies.
In conclusion, the harmonic mode of the pipe resonantly excited by a 430 Hz source is the fundamental mode (first harmonic) for a closed pipe. However, the same source will not resonate with the pipe if both ends are open due to the different nature of resonant frequencies for open pipes.
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