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COS

If R' is the resultant when the angle is 2i

R=2P cos i.
Hence by question
= 6 cos i=12 cos? -6,

2
i
12 cos?

- 6=0.

2
i 1+1+288 1+17
2

For-.]
24

24

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COS

COS

PENDULUM.

6. How many vibrations will a pendulum 39-3 inches long make in 10 minutes at a place where

g=-32.2 ?

[T=T

g Number of vibrations in 10 minutes

600_600 [g_600/32:2 x 12
T

=598:9.]
Vi

39:3

[ocr errors]
[ocr errors]

T

:. g=

دی

7. A pendulum 94 inches long makes 120 vibrations in one minute at a certain place. Find the acceleration due to gravity at that place.

39 9:87 x

12 x 4 [T=T

=32:08.]

72 (1) 8. A simple pendulum beats exactly seconds at a certain place, when the temperature is 0° C. The pendulum wire is made of platinum which expands by lot of its own length for a rise of temperature from 0° C. to 100° C. Find how much a clock regulated by this pendulum would lose per day at the same place when the temperature is 14° C.

[Let l=length of pendulum at 0° C. Then the length at 14° C.=1(1+usioo). Thus time of vibration (1+1boo)

TV1 tukioo g

T=T

[ocr errors]

=

[merged small][merged small][ocr errors][merged small]

Hence number of seconds lost per day

24 x 60 x 60 x 7

35•2.

116700 9. A pendulum which beats seconds at a certain place where g=32-2 is carried to a place where g=32.197. How much would a clock regulated by this pendulum lose or gain per day?

[It would lose 4 seconds per day.]

10. If W be the mass of the bob of a simple pendulum L the angle which the cord makes with the vertical when the pendulum is at the extremity of its range, the angle which the cord makes with the vertical in any other position of the pendulum, and T the tension of the cord, prove that

T=Wg(3 cos 0 – 2 cos L). [Let O be the point of suspension of the pendulum. A the extremity of the range, and B the point where the cord makes an angle 0 with the vertical. Let t' be the tension of the cord at the point B, if B were the extremity of its range. Then r=Wg cos 0.

The velocity acquired by the bob in falling from A to B is found from V2=2g8=2gl cos -cos L) and the centrifugal force at the point B is therefore

=W x 2g(cos 0 - cos L). The tension at the point B therefore =Wg cos A+ 2 Wg(cos - cos L)=Wg(3cos 0 – 2cos L).]

11. A simple pendulum the mass of whose bob is 100 grammes swings through an arc of 5° on each side of the position of equilibrium. Find the tension of the cord (1) when the pendulum is at the extremity of its range; (2) when the pendulum is passing through the position of equilibrium.

[(1) T=99•62 grammes weight; (2) T=100-76 grammes weight.]

12. The semivertical angle of a conical pendulum is 20° and the mass of the bob 2 kilogrammes. Find the tension of the cord. Find also the period and linear velocity of the bob, if the length of the cord is that of the second's pendulum.

[To find T', the tension of the cord, by resolving vertically we have T'cos O=W; :: T 32:13 grammes weight. To find T the period, by resolving horizontally, we have

T'sin 0= Ww-r; . Ww?l=T":

2 93 97

Wg

cos O

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T27

=2cos 0.7 =2 x .969=1.938 secs. g

9
To find V, the linear velocity, we have

g
1

sin
V2=w2p2= .72sin20=gl.
I cos 0

[ocr errors]

cos O n?cos ..

V=110.2 cms. per sec.

SIMPLE HARMONIC MOTION.

13. Define Simple Harmonic Motion. A particle is moving with simple harmonic motion over a range of 1} feet on each side of the middle position five

times per second. Find the maximum velocity and the velocity at one eighth of the period after leaving the middle position.

[If at every instant a perpendicular AB be drawn from A, which moves in a circle with uniform velocity, to a fixed diameter of the circle, then the motion of B is simple harmonic. Maximum velocity (V)=

271 273
T

=157 feet per second. Velocity at one eighth period

27_15/2

feet second.

per 8 2

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14. Define the terms Amplitude, Epoch, Period, Phase, as used in simple harmonic motion. [See Thomson & Tait, s 71.]

27 15. Prove the formula x=R sin 7t, which gives the displacement x at any time t of a point executing simple harmonic motion, of amplitude R and period T.

16. A particle weighing to lb. moves backward and forward in a straight line, 3 inches long, with simple harmonic motion 25 times per second. Find the force at the ends of the range, also the forces at a quarter and at one half the maximum distance from the centre.

[Force at ends of range=mwr kinetic units

[blocks in formation]

Force at a quarter distance from centre

= mwar

=154:22 poundals. Force at half distance from centre

=308·437 poundals.]

17. A mass of 17 of a gramme executes 512 simple harmonic periods per second, and the amplitude of its motion is of a centimetre on each side of the middle position. Find the maximum force and the force at a distance of zo of a centimetre from the middle.

[258736-128 dynes=263.64 grammes weight.
51747.2256 dynes= 52.73 grammes weight.]

CENTRIFUGAL FORCES. 18. A mass of zoth gramme revolves in a circle one-third of a centimetre in radius and makes 65 revolutions per second. Find the centrifugal force in dynes and in grammes weight.

(g=981 centimetres per second per second.)

2

2

==

(130)

dynes
60
-- 2780:05 dynes
=283 grammes weight.]

19. Two particles each a pound mass revolve in circular orbits. The period of the one is :08 of a second, and the radius of its orbit 4 inches, the period of the other is .27 of a second, and the radius of its orbit 9 inches. Find the force acting on the body

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