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Question: A pendulum clock consists of an iron rod connected to a small, heavy bob. If it is designed to keep ...

A pendulum clock consists of an iron rod connected to a small, heavy bob. If it is designed to keep correct time at 20∘C20^\circ C, how fast or slow will it go in 24 hours at 40∘C40^\circ C? Coefficient of linear expansion of iron =1.2×10−5∘C−11.2 \times {10^{-5}}\circ C^{-1}.

Explanation

Solution

A motion in which a particle undergoes periodic motion is called Simple harmonic motion (S.H.M). Not every periodic motion is S.H.M but every S.H.M is periodic motion. The revolution of earth about the sun is an example of periodic motion but it is not simple harmonic. A motion is said to be simple harmonic only if the acceleration of the particle is the function of first power of displacement and having direction opposite of the displacement. The motion of a pendulum is an example of SHM.

Formula used:
T=2πlgT = 2\pi \sqrt{\dfrac lg}, Δl=l∘αΔT\Delta l = l_\circ \alpha \Delta T

Complete step by step answer:
The motion of the pendulum of the clock is an example of simple harmonic motion since the acceleration of the pendulum is directly proportional to the first power of displacement. The time period of a simple pendulum is given by the expression:T=2πlgT = 2\pi \sqrt{\dfrac{l}{g}}.
As we can see in the expression, the time period of the simple pendulum depends upon the length of the pendulum. As we raise the temperature of the pendulum, the string will expand so the length ‘l’ will be changed. Once our length is changed, the time period will automatically be changed.
Now, as we want the change in time period, we will first of all take log on both sides and then differentiate the time period expression:
  ⟹  log(T)=log(2π)+12log(l)−12log(g)\implies log(T) = log(2\pi) + \dfrac12 log(l) - \dfrac12 log(g)
Now, dl=lαΔT=l(1.2×10−5×20)=2.4×10−4×ldl = l \alpha \Delta T = l (1.2 \times 10^{-5} \times 20) = 2.4 \times 10^{-4} \times l
Thus dll=2.4×10−4\dfrac{dl}{l} = 2.4\times 10^{-4}
Now, differentiating, we get;
dTT=dl2l=2.4×10−42=1.2×10−4\dfrac{dT}{T} = \dfrac{dl}{2l} = \dfrac{2.4 \times 10^{-4}}{2} = 1.2\times 10^{-4}
Thus, percentage change in time period:
dTT%=1.2×10−4×100=1.2×10−2\dfrac{dT}{T} \% = 1.2\times 10^{-4} \times 100 = 1.2 \times 10^{-2}%
Thus, in 24 hours, it’ll get slow by 24×1.2×10−2×60≈17 min 20 sec24\times 1.2\times 10^{-2} \times 60 \approx 17\ min\ 20\ sec

Note:
Since the length of the pendulum is increased, the time period will increase, which means the clock will take more time to cover a particular time. But in reality, the time is little less. Thus, we will say that the clock is lagging or becoming slower.