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Bukhovtsev-et-al-Problems-in-Elementary-Physics

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CHAPTER 4<br />

OSCILLATIONS<br />

AND WAVES<br />

4-1. Mechanic<strong>al</strong> Oscillations<br />

621. The vertic<strong>al</strong> component of the tension force T is equ<strong>al</strong> to F= t cos ex<br />

(Fig. 477). For the conic<strong>al</strong> pendulum F= mg, s<strong>in</strong>ce the weight has no acceleration<br />

<strong>in</strong> the vertic<strong>al</strong> plane.<br />

When the mathematic<strong>al</strong> pendulum is deflected the maximum from the<br />

position of equilibrium (through the angle a), the result<strong>in</strong>g force is directed<br />

at a tangent to the trajectory of the weight.<br />

Therefore, T = mg cos ct.<br />

When the weight is deflected through the angle a, the tension of the thread<br />

of the conic<strong>al</strong> pendulum will be greater.<br />

622. The period of oscillations of the pendulum on the surface of the Earth<br />

is To= 2n .. / I and at an <strong>al</strong>titude of h above the Earth T I = 2n .. / I<br />

JI g JI it<br />

The number of oscillations a day is N1=24X60x60 ~l ~ :1. Therefore,<br />

at an <strong>al</strong>titude of h above the Earth the clock will be slower by the time<br />

To'<br />

1it 1=N1 (T1-To)=k ( 1-T";)<br />

The ratio b<strong>et</strong>ween the periods is ;: =<br />

as follows from the law of gravitation. Hence.<br />

At kh ~ kh ~ 2 7 d<br />

u 1= R+h - R - . secon s<br />

V ~ = R:h<br />

If the. clock is lowered <strong>in</strong>to a m<strong>in</strong>e. the acceleration<br />

ti . g?, R-h. 4n RS 1 d<br />

ra 101S--g=--r-. smce g="3 p R" an g2=<br />

4n 1<br />

= l' 3 (R - h)3P(R ~ h)t (see Problem 234).<br />

Hence,<br />

;:=V~=VRR h~1-2~<br />

In this case the clock will .be slower by the time<br />

Fig. 477<br />

b-tl=k (<br />

To ) kh.- d<br />

l-r; =2R = 1.35 secon s

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