A pendulum clock is 5sec fast at temperature of ${15^0}C$ and 10sec slow at a temperature of ${30^0}C$. At temperature does it give the correct time-
(A) ${18^0}C$
(B) \[{20^0}C\]
(C) ${22.5^0}C$
(D) ${25^0}C$
Answer
601.5k+ views
Hint: This could be simply solved by applying the formula of time period. Also, the concept of linear expansion is applied.
Formula used: Here, we will use the basic formula of speed, distance and time:
$T = 2\pi \sqrt {\dfrac{l}{g}} $
Here, $T$is the time
$l$is the length
Complete step by step answer:
We will start by considering the formula:
$T = 2\pi \sqrt {\dfrac{l}{g}} $
$ \Rightarrow T \propto {l^{1/2}}$
Now from above, we can say that,
\[\dfrac{{\Delta T}}{T} = \dfrac{1}{2}\dfrac{{\Delta l}}{l}\]
And by linear expansion formula:
$\Delta l = l\alpha \Delta t$
Change in temp=final temp-initial temp
If the clock is 5 sec fast, it means that in 1 hour alarm rises 5sec before
So, change of temp=-5
Similarly, for case 1:
5 sec fast at ${15^0}C$
$\dfrac{{ - 5}}{T} \propto \Delta t = \dfrac{1}{2} \propto \left( {15 - t} \right)$
Similarly, for case 2;
$\dfrac{{10}}{T} = \dfrac{1}{2} \propto \left( {30 - t} \right)$
So, from above equations:
$ - 0.5 = \dfrac{{15 - t}}{{30 - t}}$
$ \Rightarrow 30 - t = 2\left( {t - 15} \right)$
We get temp=${20^0}C$
Thus, we need to select the correct option.
The correct option is B.
Additional Information : Simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side. The time interval of each complete vibration is the same.
Note: Some of the examples of SHM are Swings that we see in the park is an example of simple harmonic motion. The back and forth, repetitive movements of the swing against the restoring force is the simple harmonic motion.
The pendulum oscillating back and forth from the mean position is an example of simple harmonic motion.
The process of hearing is impossible without simple harmonic motion. The sound waves that enter our ear causes the eardrum to vibrate back and forth.
Formula used: Here, we will use the basic formula of speed, distance and time:
$T = 2\pi \sqrt {\dfrac{l}{g}} $
Here, $T$is the time
$l$is the length
Complete step by step answer:
We will start by considering the formula:
$T = 2\pi \sqrt {\dfrac{l}{g}} $
$ \Rightarrow T \propto {l^{1/2}}$
Now from above, we can say that,
\[\dfrac{{\Delta T}}{T} = \dfrac{1}{2}\dfrac{{\Delta l}}{l}\]
And by linear expansion formula:
$\Delta l = l\alpha \Delta t$
Change in temp=final temp-initial temp
If the clock is 5 sec fast, it means that in 1 hour alarm rises 5sec before
So, change of temp=-5
Similarly, for case 1:
5 sec fast at ${15^0}C$
$\dfrac{{ - 5}}{T} \propto \Delta t = \dfrac{1}{2} \propto \left( {15 - t} \right)$
Similarly, for case 2;
$\dfrac{{10}}{T} = \dfrac{1}{2} \propto \left( {30 - t} \right)$
So, from above equations:
$ - 0.5 = \dfrac{{15 - t}}{{30 - t}}$
$ \Rightarrow 30 - t = 2\left( {t - 15} \right)$
We get temp=${20^0}C$
Thus, we need to select the correct option.
The correct option is B.
Additional Information : Simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side. The time interval of each complete vibration is the same.
Note: Some of the examples of SHM are Swings that we see in the park is an example of simple harmonic motion. The back and forth, repetitive movements of the swing against the restoring force is the simple harmonic motion.
The pendulum oscillating back and forth from the mean position is an example of simple harmonic motion.
The process of hearing is impossible without simple harmonic motion. The sound waves that enter our ear causes the eardrum to vibrate back and forth.
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