http://vhmthphy.vhcc.edu/ > Physics I > info > syllabus etc.

http://vhmthphy.vhcc.edu/ > Physics I > Assts then go to the very bottom of the page for a link to notes.

Physics I Quiz 0825

1.  If you count 50 cycles of a pendulum in 90 seconds then

• What is the period of the pendulum, in seconds? 

The period of the pendulum is # seconds / # cycles:

period = 90 seconds / (50 cycles) = 1.8 sec / cycle, or just 1.8 sec.

• What is the frequency of the pendulum, in cycles / minute?

The period of the pendulum is 1.8 sec, so in 1 minute we will observe

# cycles in 1 minute = 60 sec / (1.8 sec) = 33.33 cycles.

So the frequency is 33.33 cycles / minute.

• What is the frequency of the pendulum, in cycles / second?

To get the number of cycles in a second we divide the number of cycles by the number of seconds:

Frequency = # of cycles / # of seconds = 50 cycles / (90 sec) = .55... cycles / sec.

2.  If your observations indicate that the period of a pendulum is 1.47 seconds, and if it is known that the theoretical period is 1.52 seconds, then

• What proportion of the theoretical period is the discrepancy?

The discrepancy is | 1.52 sec - 1.47 sec | = .05 sec.

As a proportion of the theoretical period this is .05 sec / (1.52 sec) = .033 approx..

• What is the discrepancy between these results, expressed as a percent of the theoretical period?

The percent is the proportion * 100%, or .033 * 100% = 3.3%.

3.  The theoretical period of a pendulum of length L, measured in cm, is T = .20 sqrt(L), with T in seconds.

• What is the theoretical period of a pendulum of length 50 cm?

The theoretical period is

T = .2 sqrt(50) = .2 * 71 = 1.42, approx..

meaning (since the length is in cm) that the period is 1.42 sec.

• What is the theoretical length of a pendulum whose period is .34 seconds?

T = .2 sqrt(L).  We want to find the L corresponding to a given T.  So we solve the equation for L:

Starting with

T = .2 sqrt(L), we first divide both sides by .2 to get

T / .2 = sqrt(L), then we square both sides to get

(T / .2)^2 = L.

So L = (T / .2)^2 = T^2 / (.2^2) = T^2 / (.04) = 25 T^2.

So for T = .34 sec we get

L = 25 * (.34 sec)^2 = 2.7 cm.

4.  While driving in a car you observe that in 80 complete cycles of a pendulum of length 14 cm you travel .92 mile.  What is your average speed during this time, in miles / hour?

A pendulum of length 14 cm has period T = .2 sqrt(14) = .2 ( 3.7) = .74, about .74 sec.

80 cycles will take .74 sec / cycle * 80 cycles = 59 sec, approx..

So we travel .92 miles in 59 sec.

Speed is | velocity |.

The average velocity is the average rate at which position changes:

ave velocity = change in position / change in clock time = .92 miles / (59 sec) = .0154 miles / second.

To get this in miles / hour, we multiply by the number of seconds in an hour:

(.0154 mile / second) * ( 3600 seconds / hour) = 55 miles / hour.

5.  By how much will the period of a pendulum change if its length is changed from 15 cm to 20 cm? 

• What therefore is the average rate of change of its period with respect to its length between these two lengths?

• What is the average rate of change of period with respect to length between lengths of 70 cm and 75 cm?

Suppose that between clock time 109.0 sec and 120.2 sec the number of cycles counted changes from 12 to 21.

Frequency is rate at which count changes with respect to clock time so

freq

= change in count / change in clock time

= (21 cycles - 12 cycles) / (120.2 sec - 109.0 sec)

= 9 cycles / (11.2 sec) = .80 cycles / sec.

151.3

159.2