Lab Methods - Graph

In laboratory experiments, graph are employed to determine how one measured variable depends on another. These graphs generally fall into three categories: linear, semilog (logarithmic versus linear), and log-log. Which type of graph listed in the third column below would NOT be the best for plotting data to test the relationship given in the first and second columns?

		Relation		Variables Plotted		Type of graph
A.		dN/dt ∝ e−2t		Activity vs time for a radioactive isotope		Semilog
B.		eVs = hf  − W		Stopping potential vs frequency for the photoelectric effect		Linear
C.		s ∝ t2		Distance vs time for an object undergoing constant acceleration		Log-log
D.		Vout/Vin ∝ 1/ω		Gain vs frequency for a low-pass filter		Linear
E.		P ∝ T4		Power radiated vs temperature for blackbody radiation		Log-log

(GR9677 #27)

Solution:

(A) dN/dt ∝ e^−2t
log [dN/dt] ∝ log e^−2t
log [dN/dt] ∝ −2t
Activity vs time graph has a semilog plot.

(B) V_s ∝ f
Stopping potential vs frequency graph has a linear plot.

(C) s ∝ t²
log s ∝ 2 log t
Distance vs time graph has a log-log plot.

(D) V_out/V_in ∝ 1/ω
V_out/V_in ∝ ω⁻¹
log [V_out/V_in] ∝ − log ω
Gain vs frequency graph has a log-log plot, NOT linear.

(E) P ∝ T⁴
log P ∝ 4 log T
Power vs temperature graph has a log-log plot.

Answer: D

Notes:

A semilogarithmic graph has one axis in logarithmic scale and the other in linear scale.

A log-log graphs has both axes in logarithmic scale.

Electromagnetism - Oscilloscope

The figure above represents the trace on the screen of cathode ray oscilloscope. The screen is graduated in centimeters. The spot on the screen moves horizontally with a constant speed of 0.5 centimeter/millisecond and the vertical scale is 2 volts/centimeter. The signal is a superposition of two oscillations. Which of the following are most nearly the observed amplitude and frequency of these two oscillations? 

Oscillation 1 Oscillation 2 A. 5V, 250Hz 2.5V, 1000Hz B.  1.5V, 250Hz 3V, 1500Hz  C. 5V, 6Hz 2V, 2Hz  D. 2.5V, 83Hz 1.25V, 500Hz  E. 6.14V, 98Hz                1.35V, 257Hz

(GR9677 #28)

Solution:

The graph shows one big λ consists of 6 small λs.
λ_big ≈ 6 cm
λ_small ≈ 1 cm 

v = λf

Given: v = 0.5 cm/ms
f_big = v/λ_b = 0.5 cm/(6 cm ms) = 1/ (12 ms) = 10³/(12 s) = 83 Hz (Osc. 1)
f_small = v/λ_s = 0.5 cm/(1 cm ms) = 1/ (2 ms) = 10³/(2 s) = 500 Hz (Osc. 2)

Answer: D

Lab Methods - Voltage Amplifier

In a voltage amplifier, which of the following is NOT usually a result of introducing negative feedback?

A. Increased amplification
B. Increased bandwidth
C. Increased stability
D. Decreased distortion
E. Decreased voltage gain

(GR9677 #72)

Solution:

Negative feedback → noise cancellation.
Answer B, C, D, and E are benefits of negative feedback.

Positive feedback → increased amplification.

Answer: A

Lab Methods - Oscilloscope

The outputs of two electrical oscillators are compared on an oscilloscope screen. The oscilloscope spot is initially at the center of the screen. Oscillator Y is connected to the vertical terminals of the oscilloscope and oscillator X to the horizontal terminals. Which of the following patterns could appear on the oscilloscope screen, if the frequency of oscillator Y is twice that of oscillator X?

(GR9277 #17)

Solution:

→

Answer:

Lab Methods - Standard Deviation

The magnitude of the force F on an object can be determined by measuring both the mass m of an object and the magnitude of its acceleration a, where F = ma. Assume that these measurements are uncorrelated and normally distributed. if the standard deviations of the measurements of the mass and acceleration are σ_m and σ_a respectively, then σ_F/F is

A.

B.

C.

D.

E.

(GR9277 #48)

Solution:

Multiplying two quantities with uncertainty:

u = x ⋅ y



For F = ma:



Answer: C

Electromagnetism - Oscilloscope

The circuit shown above is used to measure the size of the capacitance C. The y-coordinate of the spot on the oscilloscope screen is proportional to the potential difference across R, and the x-coordinate of the spot is swept at a constant speed s. The switch is closed and then opened. One can then calculate C from the shape and the size of the curve on the screen plus a knowledge of which of the following?

A. V₀ and R
B. s and R
C. s and V₀
D. R and R'
E. The sensitivity of the oscilloscope

(GR9277 #86)

Solution:

The voltage of a capacitor follows an exponential decay:

V(t) = V₀e^−t/RC 

C depends on V(t), V₀, t, and R.

V₀ is given and known from the beginning of measurement.

The y-coordinate of the spot on the oscilloscope screen is proportional to the potential difference across R. Thus, V(t) is the shape and the size of the curve on the screen.
 
The x-coordinate of the spot is swept at a constant speed s. From this we can calculate time t.

Therefore, we can calculate C from the shape and the size of the curve plus a knowledge of s and R.

Answer: B

Lab Methods - Graph

The gain of an amplifier is plotted versus angular frequency ω in the diagram above. If K and a are positive constants, the frequency dependence of the gain near ω = 3 × 10⁵ second⁻¹ is most accurately expressed by

A. Ke^−aω
B. Kω²
C. Kω
D. Kω⁻¹
E. Kω⁻²

(GR8677 #39)

Solution:

(A) g = Ke^−aω → decreasing (exponential decay). FALSE.


(B) g = Kω² → increasing (quadratic function/parabola). FALSE.


(C) g = Kω → increasing (linear function). FALSE.
(D) g = K/ω → decreasing (linear function). FALSE.

(E) g = K/ω² → decreasing (quadratic function). TRUE.

Check:
At ω = 10⁶, g ≈ 10
g = K/ω² 
10 = K/10¹²
K = 10¹³

At ω = 3 × 10⁵, K = 10¹³
g = 10¹³/(9 × 10¹⁰) ≈ 0.11 × 10³
10²  g  10³

Answer: E

Lab Methods - Precision

Five classes of students measure the height of a building. Each classes uses a different method and each measures the height many different times. The data for each class are plotted below. Which class made the most precise measurement?

(GR0177 #15)

Solution:

The accuracy is how close the peak is to the reference value.

The precision is how narrow the peak is.

So we look for the graph with the narrowest peak.

Answer: A

Lab Methods - Uncertainty

A student makes 10 one-second measurement of the disintegration of a sample of a long lived radioactive isotope and obtains a following values: 3, 0, 2, 1, 2, 4, 0, 1, 2, 5. How long should the student count to establish the rate to an uncertainty of 1 percent?

A. 80 s
B. 160 s
C. 2000 s
D. 5000 s
E. 6400 s

(GR0177 #16)

Solution:

Radioactive decay can be described by Poisson Distribution.

Poisson Distribution (PD):
Probability distribution of discrete events over an interval (time. distance, etc)

In PD, Standard Deviation, σ = √μ  (see problem GR8677 #40)
μ = λT = expected value
λ = average rate
T = time interval

% Uncertainty = (σ/μ) × 100%

σ/μ = 0.01
√μ/μ = 10⁻²
μ/μ² = 10⁻⁴
1/μ = 1/10⁴
μ = λT = 10⁴

λ = (3 + 0 + 2 + 1 + 2 + 4 + 0 + 1 + 2 + 5)/10 = 2
T = 10⁴/2 = 5000

Answer: D