p19_023.pdf

23. We note that if the pendulum shortens, its frequency of oscillation will increase, thereby causing it

to record more units of time (“ticks”) than have actually passed during an interval. Thus, as the

pendulum contracts (this problem involves cooling the brass wire), the pendulum will “run fast.” Since

the “direction” of the error has now been discussed, the remaining calculations are understood to be in

absolute value. The diﬀerential of the equation for the pendulum period in Chapter 16 is

√ gL

which we divide by the period equation T =2 π L/g (and replace diﬀerentials with

dT = 1

2 (2 π ) dL

∆

’s) to obtain

∆ T

= 1

2 |

∆ L

= 1

∆ T

where we use Eq. 19-9 (in absolute value) in the last step. Thus, the (unitless) fractional change in

period is

∆ T

2 19

10 − 6 / C ◦ (20C ◦ )=1 . 9

10 − 4

using Table 19-2. We can express this in “mixed units” fashion by recalling that there are 3600 s in an

hour. Thus, (3600s / h)(1 . 9

10 − 4 )=0 . 68 s/h.

2 α

= 1