Q16P - Fundamentals Of Physics

Question

Time standards are now based on atomic clocks. A promising second standard is based on pulsars, which are rotating neutron stars (highly compact stars consisting only of neutrons). Some rotate at a rate that is highly stable, sending out a radio beacon that sweeps briefly across Earth once with each rotation, like a lighthouse beacon. Pulsar PSR 1937 + 21 is an example; it rotates once every 1.557 806 448 872 75 $\pm$ 3 ms, where the trailing $\pm$3 indicates the uncertainty in the last decimal place (it does not mean $\pm$3 ms). (a) How many rotations does PSR 1937 + 21 make in 7.00 days? (b) How much time does the pulsar take to rotate exactly one million times and (c) what is the associated uncertainty?

Step-by-Step Solution

Answer

(a) Number of rotations in 7.00 days is $3.88\times {10}^8$

(b) Time taken by the Pulsar for 1 million rotations is $1557.80644887275 s$

(c) Associated uncertainty is $\pm 3\times {10}^{-11} \text{s}$

Step 1: Given data

The time period of rotation of the Pulsar, ${\text{P}}_t=5578064487275 ms$

The uncertainty per rotation is $\pm 3\times {10}^{-17}s$

Step 2: Understanding the atomic clock

Pulsars are rapidly rotating stars that emit radio pulses with a very regular frequency. Atomic clocks measure time by measuring the frequency of radiation emitted by atoms. An atom's natural oscillations work like the pendulum in a grandfather clock.

Step 3: (a) Determination of the number of rotations in 7 days

The frequency is reciprocal of time. Therefore,

$\text{f=}\frac{1 rotation}{1.55780644887275\times {10}^{-3} s}$ … (i)

Multiply this frequency by the given time 7days in seconds to get the number of rotations. 

Thus, the Pulsar makes $3.88\times {10}^8$ rotations in 7 days.

Step 3:(b) Determination of the time taken by Pulsar for 1 million rotations

Using equation (i), time is calculated as: 

$n=f\times t$

Here number of rotations is $1\times {10}^6$. So

Thus, the time taken by the Pulsar to rotate one million times is $\text{1557.80644887275 s}$

Step 3:(c) Determination of the uncertainty

The time uncertainty per revolution is $\pm 3\times {10}^{-17}s$

 

For one million revolutions, associated uncertainty is,

  $$\begin{gathered} \text{Associated Uncertainty}=\pm 3\times {10}^{-17}\times 1\times {10}^6 \\ =\pm 3\times {10}^{-11}\text{s} \end{gathered}$$

Thus, the associated uncertainty is.${\text{±3×10}}^{-11}s$