Standing Waves

Objective

Determine the right relationship between spring length, velocity to create a standing wave with different frequency.

Procedure

SET UP THE FREQUENCY GENERATOR

SET UP THE MASS

OBSERVE THE STANDING WAVE

 Data

		Length (m)	Mass (kg)	Density (μ, kg/m)
	String	1.98	0.00061	0.0003087
	Case #1	Length (L)	Hanging mass (kg)		Tension	Wave Speed
		1.45	0.20		1.96	79.72
		# of wave (n)	Frequency (Hz)	wavelength (λ)	1/λ
		1	28	2.90	0.34
		2	56	1.45	0.69
		3	83	0.97	1.03
		4	111	0.73	1.38
		5	138	0.58	1.72
		6	167	0.48	2.07
		7	194	0.41	2.41
		8	222	0.36	2.76
		9	249	0.32	3.10
		10	277	0.29	3.45
	Case #2	Length (L)	Hanging mass (kg)		Tension	Wave Speed
		1.45	0.050		0.49	39.86
		# of wave (n)	Frequency (Hz)	Wavelength (λ)	1/λ
		1	14	2.90	0.34
		2	28	1.45	0.69
		3	39	0.97	1.03
		4	47	0.73	1.38
		5	70	0.58	1.72
		6	84	0.48	2.07

 CASE 1

Graph 1/lambda vs f

the slope is the wave speed

CASE 2

Graph 1/lambda vs f

the slope is the wave speed

The ratio of the frequencies to same harmonics:

Conclusion
Since v is proportional to tension, we made the tension in case 2 four times bigger than case 1, so we should get the wave speed of case 2 two times faster than case 1. The wave speed of case 2 was about 2.06 times faster than case 1, which is what we expected. And we also got the same ratio of frequency in different modes. This experiment was very successful.

Waves

Objective
Find the relationship between frequency and wavelength.

Procedure

 

 

Data

 

 

 

f*(lambda)=5.79683, 5.46612, 4.7286 (m/s)

 

Conclusion

 

Since v=sqrt(Tension/linear density)

When the distance between two people gets longer, the tension increase and the density decrease.

Therefore, the velocity increase.

 

=> when the tension (density) is about the same, we have f*(lambda)=constant=velocity.

 

 



 

 

Fluid Dynamics

Objective: Verify Bernoulli's Equation-- P+Dgh+1/2Dv^2=constant

When delta P=0, V=sqrt(2gh)

Flow rate = Volume per seconds = speed * area
Therefore, A=Volume/(speed * time)
                     =V/sqrt(2gh)*t

We measure the Volume, time, and water depth to find the area of the little hole and verify the Bernoulli’s Equation.

Measure the hight

Measure the time

 

Measure the area of the hole

Compare to the theoretical Area

 

Test #	1	2	3	4	5	6
Time (s)	8.60	8.25	8.78	8.33	8.78	8.40

Run	Time to empty (s)	Theoretical Time (s)	% Error
1	8.60	7.01	22.68%
2	8.25		17.69%
3	8.78		25.25%
4	8.33		18.83%
5	8.78		25.25%
6	8.40		19.83%

  
Measured diameter: 6.0mm
calculated diameter: 5.3

Percent Error = 13.2%

Conclusion
There is about 20% error on time and 13% error on the area of the hole. We used v=sqrt(2gh) to find the time; however, the water depth h is changing during the process. I think the error is mainly caused by the changing of h. Also, our reaction time also effect the result, and the hole might not be a circle so A is not equal to [pi]r^2. Other than those, we verify the Bernoulli's equation.

Fluid Statics

Objective: Experience and calculate the buoyant force.

Part A
Underwater Weighting Method

Measure the weight of the cylinder---1.1023N
Attach the metal cylinder to a string and put it in the water.
Measure the reading on the string---0.73N
The difference is the buoyant force.
B=W-T
   =0.3723497 N
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Part B
Displaced Fluid Method

Fill up the beaker with water and measure the total mass---0.17872g
Put the metal cylinder into the beaker and measure the exceeded water---0.0316g
Buoyant force equals to the weight of displaced water.
B=W=0.0316g * 9.8 m/s^2
   =0.30968 N
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Part C
Volume of Object Method

Use the caliper to measure the diameter and height of the cylinder
d=0.253m, h=0.0771m
V=pi*r^2*h
   =3.876*10^-5
B=d*V*g
   =1000*3.876^-5*9.8
   =0.379894 N
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Question:
1. The error of Part A mainly come from the balance and the reading on the string. It's basically a very accurate way to measure the buoyant force. For, Part B, we use the concept of Archimede's principle. Since the volume of the cylinder is small and the exceeded water was just about 30-40 ml, the error of this method could be very large. Any lost of 1 ml costs about 2.5 percent error. Therefore, Part B is not a good way to determine the buoyant force. For Part C, the errors come from (1) the volume of the cylinder. (2) the density of the water. It is not easy to measure the true values of the diameter and height by using a caliper. And the density of water at the room temperature might not be exactly 1000kg/m^3. However, Part C is still a good way to find the buoyant force even it is not as accurate as Part A.

2. Part A (Underwater weighting method ) is the most accurate.

3. If ,in Part A, the cylinder had been touching the bottom of the water container, the container would exert a force N to the cylinder. Then W=T+B+N or B=W-T-N. The value would be to low comparing to B=W-T.