Question

3: A simple pendulum with a period of 2 s is used in a clock. The “negligible” bar connecting the pendulum bob to the pivot is made of aluminum (α = 2.4x10^-5/◦C)The clock was built to keep accurate time at 24 ◦C. If the clock is used at an average temperature of 35 ◦C, what correction is needed after 30 days? It will be useful to keep every digit you can throughout the calculation since the differences are quite small. I used 9.81 m/s2 for the acceleration due to gravity.

Answer is 338s. Show me all the work please!

Answer #1

Coefficient of linear expansion = 2.4 x10
^{-5} / ^{o} C

Period T = 2 s

Length of the pendulam L = L

We know T = 2[L/g]

T ^{2} = 4^{2}
(L/g)

From this L = g [T ^{2}/4^{2} )

= 9.81 [2 ^{2} /(4x^{2} ) ]

= 9.81 /^{2}

= 0.9939 m

Length at 35 ^{o} C is L ' = L [1+(t '- t ) ]

= 0.9930 [1+(2.4x10 ^{-5})(35-24)]

= 0.99422 m

Period at 35 ^{o} C is T '= 2[L ' / g
]

= 2(22/7) [0.99422/9.81]

= 2.0002639 s

30 days = 30 x 24 hrs = 30 x24 x 3600 s

Number of oscillations at 24 ^{o} C in 30 days , N = (30
x24 x3600) / 2 s

Correction = N(2.0002639 -2)

= 342.12 s

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