Lab 11 Bulk Modulus PDF

Preview

Summary

Physics 151L Spring 2019: Lab #11: Bulk Modulus ...

Description

Bulk Modulus of Air Objective: To calculate the bulk modulus of air (B) using two different methods, the static method which calculates Bs as the transfer of initial energy in the form of pressure exerted from air compression, based off of the drop distance of the ball (h) and the dynamic method which calculates Bd based off of the average period of oscillation (T2). Theory: Experimental setup:

Diagram 1: The experimental setup of the bottle apparatus used to measure the bulk modulus of air using the dynamic and static method.

The objective of this experiment is to measure the bulk modulus of air (B) which is observed with the force (F) exerted under a certain pressure (P) in a certain area (A). The equation F=

P demonstrates the general relationship of force and pressure, which is changed A

correspondingly when the bulk modulus changes the volume of the system, The bulk modulus can be theoretically calculated from

Δ P=−B

ΔV . V

B=γP which is derived from the

previous equation. During the experiment, a ball will be dropped in a glass tube to measure the force exerted by the gas compressed in the bottle system according to the bulk modulus. Since the bulk modulus acts upon the change in volume the change in displacement of the ball

ΔV ,

it has a corresponding relationship to

Δ x , which can be calculated using the equation

=A Δ x . The relationship between the change in force

ΔV

Δ F the change in displacement of

the ball

Δx,

is calculated using

Δ F=

−A 2 B Δ x . The spring constant, keff, V

2

A B V

can

be plugged into the equations used for the static and dynamic methods used to calculate the experimental bulk modulus. For the dynamic method, the oscillations of the ball when dropped in the tube will be measured as periods (T2) calculated as

B=

4 π 2 mV . The static method A2 T 2

measures the transfer of initial energy into final energy of the maximum air compression exerted onto the ball at a particular drop point, which will exert force back onto the ball causing it to oscillate back up the tube. This relationship between the drop point of the ball and the bulk modulus for the static method will be calculated as

B=

2mgV . h A2

Procedure: 1. Measure the experimental constants for the experiment: mass of the ball (kg), the radius of the glass tube (m), and the volume of the bottle (m3) 2. Calculate the cross sectional area (A) using the radius. 3. Assemble the bottle on table according to diagram 1 Dynamic Method: 4. Obtain a stopwatch to record the period of three oscillations. 5. Drop the ball in the glass tube, stating the time at the first drop then stopping the time at the third drop distance. 6. Record the data on table 1 for each trial (3T) 7. Repeat for 5 trials total. 8. Take each trial’s value for 3T and divide it by 3 to get the period for each trial (T) 9. Calculate the average T, then the error sdm(T) using the data for trials 1-5. 10. Calculate the bulk modulus Bd Static Method: 11. Drop the ball in the glass tube and record the drop distance (h) for each trial on table 2. 12. Repeat for 5 trials total 13. Calculate the average h, then the error sdm(h) using the data for trials 1-5. 14. Calculate the bulk modulus Bs

Data: Experimental constants: Mass of ball: 0.00818 kg Radius of tube: 0.00635 m Volume of bottle: 0.00388 m3 Pressure of the classroom (P): 101400 PA

Table 1: Experimental data and calculations of the Dynamic method

Table 2: Experimental data and calculations of the Static method Calculations: Experimental Calculations: 2 A=π R ❑ 2

0.00635 m ¿ =0.000127 m A=π ¿ B static=

2 mg h 2 A h m )(0.2716 m) 2 =1.00 x 107 Pa s 2 2 (0.000126613 m ) (0.2716 m )

(2)(0.00818 kg)(9.81 B static= B dynamic =

4 πV m 2 2 T A

3 0.00388 m ¿ ¿ = 9.79 x 104 Pa (4 π )¿ B dynamic =¿ B t h eo =γP

7 B= × 101400=141960 Pa or 1.4196 x 105 Pa 5 Error Propagation: Dynamic Method: δ Bdynamic =(2

δt )(B dynamic ) t

δ Bdynamic =(2

0.003753516869 )( 9.79× 1 04 Pa )=1.4582 × 10 3 Pa 0.504

Static Method: δ Bstatic =(

δh )(Bstatic ) h

δ Bstatic =(

0.0005403702434 )(1.00 x 1 07 Pa)=1.9895 × 1 04 Pa 0.2716

Error Discussion: ∈result |errorresult |∗100 %

p=

Confidence:

Agreement: s=

h eoretical |ResultError−T∈result |

Dynamic method:

| |

p=

s=

|

(0.00375) ∗100 % = 3.83 ¿ 10−6 % ≈ 0.0000000383 % 4 (9.79 x 1 0 Pa)

|

(9.79 x 1 0 4 Pa)−(1.4196 x 1 05 Pa ) (0.00375)

Static method:

= 117493333.3 ≈ 1.17∗1 0 8

| |

p=

s=

|

(0.000540) ∗100 % =2.714 ¿ 10−6 % ≈ 0.0000000271 % 4 (1.9895 × 10 Pa)

(1.00 x 1 07 Pa)−(1.4196 x 1 05 Pa) (0.000540)

|

= 1.824312149∗1 010 ≈ 1.82 ¿ 10 10

Results: Calculated according to the static method data obtained (see pg 3), the static method bulk modulus Bs is 1.00 x 107±δ. 1.9895× 1 04 Pa . According to the dynamic method data, the dynamic method bulk modulus Bd is 9.79 x 104 ±δ 1.4582 × 10 3 Pa . These experimental values were then compared to the theoretical bulk modulus, Btheo=1.4196 x 105

Pa to assess any

errors that could have occurred in the experiment. Based off the static and dynamic experimental and theoretical data, error propagation was calculated to determine the agreement and confidence levels for the results. Starting with the dynamic method results, the agreement value calculated, s= 1.17∗10 8 , showed that since it was greater than 1, our experimental Bd agreed with the theoretical Btheo. The confidence level, p= 0.0000000383 % , indicated that random error in our experiment was significantly underestimated since p...