A pendulum has an oscillation frequency (T) which is assumed to depend upon its length (L), load mass (m) and the acceleration of gravity (g). Determine the relationship between oscillation frequency, length, load mass and acceleration of gravity. Differentiate as well which variable does not affect the oscillation frequency.
Answers
Answer:
Mass does not affect oscillation frequency.
Explanation:
Let the bob of the pendulum makes a small angular displacement θ. When the pendulum is displaced from the equilibrium position, a restoring force tries to act upon it and it tries to bring the pendulum back to its equilibrium position. Let this restoring force be F.
Therefore, F = -mgsinθ
Now for pendulum, for small angle of θ,
sinθ[tex]\simeq[/tex]θ
Therefore, F = -mgθ
Now from Newton's 2nd law of motion,
F = m.a = -mgθ
[tex]\Rightarrow m.\frac{d^{2}x}{dt^{2}} = - mg\Theta[/tex]
Now since, x = θ.L
[tex]\Rightarrow L.\frac{d^{2}\Theta }{dt^{2}}= -g\Theta[/tex]
[tex]\Rightarrow \frac{d^{2}\Theta }{dt^{2}}= -\frac{g}{L}.\Theta[/tex]
[tex]\Rightarrow \frac{d^{2}\Theta }{dt^{2}}+\frac{g}{L}.\Theta =0[/tex]
Therefore, angular frequency
[tex]\omega ^{2}[/tex] = [tex]\frac{g}{L}[/tex]
ω = [tex]\sqrt{\frac{g}{L}}[/tex]
Also we know angular frequency is , ω = 2.π.f
where f is frequency
Therefore
2πf = [tex]\sqrt{\frac{g}{L}}[/tex]
f = [tex]\frac{1}{2 \pi }\sqrt{\frac{g}{L}}[/tex]
So from here we can see that frequency,f is independent of mass, hence it does not affect frequency.
Related Questions
A fluid with a relative density of 0.9 flows in a pipe which is 12 m long and lies at an angle of 60° to the horizontal At the top, the pipe has a diameter of 30 mm and a pressure gauge indicates a pressure of 860 kPa. At the bottom the diameter is 85 mm and a pressure gauge reading is 1 MPa. Assume the losses are negligible and determine the flov rate. Does the flow direction matter?
Answers
Answer:
[tex]Q=7.3\times 10^{-3} m^3/s[/tex]
Explanation:
Given that
At top[tex]d_2=30 mm,P_2=860 KPa ,P_1=1000 KPa,d_1=85 mm[/tex]
[tex]\rho =900\dfrac{Kg}{m^3}[/tex]
We know that
[tex]\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2[/tex]
[tex]A_1V_1=A_2V_2[/tex]
[tex]\frac{V_1}{V_2}=\left(\dfrac{d_2}{d_1}\right)^2[/tex]
[tex]\frac{V_1}{V_2}=\left(\dfrac{30}{85}\right)^2[/tex]
[tex]V_2=8.02V_1[/tex]
[tex]Z_2=12 sin60^{\circ}[/tex]
[tex]\dfrac{1000\times 1000}{900\times 9.81}+\dfrac{V_1^2}{2\times 9.81}+0=\dfrac{860\times 1000}{900\times 9.81 }+\dfrac{V_2^2}{2\times 9.81}+12 sin60^{\circ}[/tex]
So [tex]V_1=1.30[/tex]m/s
We know that flow rate Q=AV
[tex]Q=A_1V_1[/tex]
By putting the values
[tex]A_1=\dfrac{\pi}{4}d^2[/tex]
[tex]Q=7.3\times 10^{-3} m^3/s[/tex]
To find the flow rate we do not need the direction of flow,because we are just doing balancing of energy at inlet and at the exits of pipe.
A 5Kw solar system may produce enough energy to power your home. a)-True b)- False
Answers
Answer: True
Explanation:
Yes, it is true that 5 Kw solar system may produce enough energy to power your home as, on an average good quality of 5 KW solar system can produced 22 units per day enough to power all home appliances. As, a 5 KW solar system produced energy is basically depends on the three main factor that are:
Quality of the solar panel system.Location from where the solar system generated its energy.And also on the positioning of the solar system.Heat in the amount of 100 KJ is transferred directly from a hot reservoir at 1200 K to a cold reservoir at 600K.Calculate the entropy change of the two reservoirs and determine if the increase of entropy principle is satisfied.
Answers
Answer:
0.0833 k J/k
Explanation:
Given data in question
total amount of heat transfedded (Q) = 100 KJ
hot reservoir temperature R(h) = 1200 K
cold reservoir temperature R(c) = 600 k
Solution
we will apply here change of entropy (Δs) formula
Δs = [tex]\frac{Q}{R(h)}+\frac{Q}{R(c)}[/tex]
Δs = [tex]\frac{-100}{1200}+\frac{100}{600)}[/tex]
Δs = [tex]\frac{1}{12}[/tex]
Δs = 0.0833 K J/k
this change of entropy Δs is positive so we can say it is feasible and
increase of entropy principle is satisfied
Answer:
0.0837 kJ/K
Explanation:
Given:
Temperature of the cold reservoir T,cold = 600 K
Temperature of the hot reservoir T,hot= 1200 K
Heat transferred , Q=100 kJ
Now the entropy change for the cold reservoir
[tex]\bigtriangleup S,cold=-\frac{Q}{T,cold}[/tex]
[tex]\bigtriangleup S,cold=-\frac{-100}{600}[/tex]
[tex]\bigtriangleup S,cold=0.1667 kJ/K[/tex]
Now the entropy change for the cold reservoir
[tex]\bigtriangleup S,hot=-\frac{Q}{T,hot}[/tex]
[tex]\bigtriangleup S,hot=-\frac{100}{600}[/tex]
[tex]\bigtriangleup S,hot=-0.0833 kJ/K[/tex]
Therefore, the total entropy change for the two reservoir is
[tex]\bigtriangleup S=\bigtriangleup S,hot +\bigtriangleup S,cold[/tex]
thus,
ΔS=0.1667-0.0833
ΔS=0.0833 kJ/K
Since, the change of entropy is positive thus we can say it is possible and
increase of entropy principle is satisfied
Water flows in a pipe of diameter 0.5 m. The dianeter of the to 1,0 m. A U-tube manometer is of the enlargement with joining ercury levels 5 mm. Determine the flow rate as well as the pressure 3 Water pipe suddenly enlarges connected to either side pipes which contain water. The difference in m head loss as a result of the enlargement.
Answers
Answer:
Q = 0.943[tex]m^{3}/s[/tex]
[tex]h_{L}[/tex] = 0.6605 m
Explanation:
Given :
Diameter, d₁ = 0.5 m
Area, A₁ = [tex]\frac{\pi }{4}\times 0.5^{2}[/tex]
= 0.19625 [tex]m^{2}[/tex]
Enlargement diameter, d₂ = 1 m
Enlargement Area, A₂ = [tex]\frac{\pi }{4}\times 1^{2}[/tex]
= 0.785 [tex]m^{2}[/tex]
Manometric difference, h = 35 mm
=35 X [tex]10^{-3}[/tex] m
From manometer , we get
[tex]P_{1}+\rho _{w}.g.z_{1}+\rho _{m}.g.h=P_{2}+\rho _{w}.g.(z_{1}+h)[/tex]
[tex]P_{1}-P_{2}=(\rho _{w}-\rho _{m}).g.h[/tex]
[tex]\frac{P_{1}-P_{2}}{\rho _{w}.g}=(1-\frac{\rho _{m}}{\rho _{w}})\times h[/tex]
= [tex](1-13.6)\times 35\times 10^{-3}[/tex]
= -0.441
Now from newtons first law,
[tex]\frac{P_{1}-P_{2}}{\rho _{w}.g}=\frac{V_{2}^{2}-V_{1}V_{2}}{g}[/tex]
-0.441 = [tex]\frac{Q^{2}}{9.81}\times (\frac{1}{A_{2}^{2}}-\frac{1}{A_{1}A_{2}})[/tex]
-0.441 = [tex]\frac{Q^{2}}{9.81}\times (\frac{1}{0.785^{2}}-\frac{1}{(0.19625\times 0.785)^{2}})[/tex]
Therefore. Q = 0.943 [tex]m^{3} /s[/tex]
Now V₁ = [tex]\frac{Q}{A_{1}}[/tex]
= [tex]\frac{0.943}{0.19625}[/tex]
= 4.80 m/s
V₂ = [tex]\frac{Q}{A_{2}}[/tex]
= [tex]\frac{0.943}{0.785}[/tex]
= 1.20 m/s
Therefore, heat loss due to sudden enlargement is given by
[tex]h_{L}=\frac{(V_{1}-V_{2})^{2}}{2g}[/tex]
[tex]h_{L}=\frac{(4.80-1.20)^{2}}{2\times 9.81}[/tex]
= 0.6605 m
What is pre-flush and post flush in petroleum engineering?
Answers
Answer:
Pre-Flush:
It is also known as In-line Equalization. In this stage of flow equalization, all the flow passes through the equalization basin. It helps in reduction of fluctuation in pollutants concentration and flow rate and helps to control short term surges with the use of basin.
Post-Flush:
Another name for this stage is Off-line Equalization. In this stage, only overflow above a predetermined standard is diverted into the basin. It helps in reducing the fluctuations in loading by a considerable amount and helps to reduce the pumping requirement. It is basically used to capture "first flush" from combined collection systems.
Answer:
Answer:
Pre-Flush:
It is also known as In-line Equalization. In this stage of flow equalization, all the flow passes through the equalization basin. It helps in reduction of fluctuation in pollutants concentration and flow rate and helps to control short term surges with the use of basin.
Post-Flush:
Another name for this stage is Off-line Equalization. In this stage, only overflow above a predetermined standard is diverted into the basin. It helps in reducing the fluctuations in loading by a considerable amount and helps to reduce the pumping requirement. It is basically used to capture "first flush" from combined collection systems.
Explanation:
What are the air-standard assumptions?
Answers
Answer:
The air-standard assumptions are:
The working fluid is air assumed to be perfect and it behave as an ideal gas .All process are internally reversible.The cycle is modeled as closed cycle with air cooled in the chiller heat exchanger and then re-circulated to the compressor. To avoid the complications, the combustion container are replaced by combustion heat exchanger .
Three point bending is better than tensile for evaluating the strength of ceramics. a)-True b)- False
Answers
Answer:
a)-True
Explanation:
Three point bending is better than tensile for evaluating the strength of ceramics. It is got a positive benefit to tensile for evaluating the strength of ceramics.
________is defined to be the probability that a failed component or system will be restored to a repaired specified condition within a period of time when maintenance is performed in accordance with prescribed procedures
Answers
Answer:
Repairable component
Explanation:
Repairable component is defined to be the probability that a failed component or system will be restored to a repaired specified condition within a period of time when maintenance is performed in accordance with prescribed procedures.
________is defined to be the probability that a failed component or system will be restored to a repaired specified condition within a period of time when maintenance is performed in accordance with prescribed procedures
Repairable component
Convert 25 mm into in.
Answers
Answer:
25 mm = 0.984252 inches
Explanation:
Millimeter and inches are both units of distance. The conversion of millimeter into inches is shown below:
1 mm = 1/25.4 inches
From the question, we have to convert 25 mm into inches
Thus,
25 mm = (1/25.4)*25 inches
So,
[tex]25 mm=\frac{25}{25.4} inches[/tex]
Thus, solving we get:
25 mm = 0.984252 inches
What are the x and y coordinates of the centroid of the area?
Answers
Answer:
[tex]\\X_{C.G}=\frac{\int xdA}{A}\\Y_{C.G}=\frac{\int ydA}{A}[/tex]
Explanation:
The x co-ordinate of the centroid is given by:
[tex]x_{C.G}=\frac{\int xdA}{A}[/tex]
The y co-ordinate of the centroid is given by:
[tex]Y_{C.G}=\frac{\int ydA}{A}[/tex]
where
[tex]x^{}[/tex] is the x co-ordinate of a diffrential area [tex]dA[/tex]
and [tex]y^{}[/tex] is the x co-ordinate of a diffrential area [tex]dA[/tex]
See attached figure