What is a seconds pendulum how will its time period change if it is taken to the moon where the acceleration is g 6?

Text Solution

6 times12 times`(1)/(6)` times`(1)/(12)` times

Answer : C

Solution : (c) For second's pendulum at the surface of earth `implies 2=2pisqrt((l_(e))/(g_(e)))" "…(i)` For second's pendulum at the surface of moon `=2pisqrt((l_(m))/(g_(m))) " "…(ii)` From Eqs. (i) and (ii), `(l_(e))/(g_(e))=(l_(m))/(g_(m))impliesl_(m)=((g_(m))/(g_(e)))l_(e)` As `g_(m)=(1)/(6)g_(e)impliesl_(m)=(l_(e))/(6)=(1)/(6)` times (`because l_(e)=1 m`)

A second’s pendulum is taken on the surface of the moon where the acceleration due to gravity is l/6th of that of earth. Will the time period of pendulum remain the same or increase or decrease? Give a reason.

We know the time period of a simple pendulum is inversely proportional to the square root of acceleration due to gravity

i.e. T α `1/sqrt("g")`

As acceleration due to gravity on the surface of the moon decreases as compared to that of the earth
∵ gmoon < gearth
∴ Tmoon > Tearth

⇒ The time period of the second’s pendulum increases when it is taken to the surface of the moon.

Concept: Simple Pendulum for Time

Is there an error in this question or solution?

Answer

Verified

Hint: In order to solve this question, firstly we will write out the equation for the time period on the earth, then again for the time period on the moon earth by using the formula of time period at the surface of moon i.e. $T = 2\pi \sqrt {\dfrac{L}{g}} $.

Complete step-by-step solution -

We know that a seconds pendulum is a pendulum which has a time period of 2 seconds; one second for a swing in one direction and one second for the return swing with a frequency of $\dfrac{1}{2}Hz$.On earth, the time period will be-$2 = 2\pi \sqrt {\dfrac{l}{g}} .............\left( 1 \right)$On the surface of the moon, acceleration due to gravity is $\dfrac{g}{6}$.( value of g at moon is$\dfrac{1}{6}$ of value of g at earth’s surface)On the surface of the moon, the time period of a second’s pendulum will be$2\pi \sqrt {\dfrac{l}{{\dfrac{g}{6}}}} .............\left( 2 \right)$Dividing the above equations 1 and 2,We get-$\dfrac{{2}}{T_{moon}}$= $\dfrac {{2\pi \sqrt {\dfrac{{l}}{g}}}} {2\pi \sqrt {\dfrac{{l}}{\dfrac{{g}}{6} }}}$${T_{moon}} = 2\sqrt 6 s$ Therefore, the time period of a second’s pendulum on the surface of the moon will be $2\sqrt 6 s$.Hence, option C is correct.Note- While solving this question, we must know that a pendulum with any other value of period cannot be called a seconds pendulum. Thus a seconds pendulum anywhere has a period of 2 seconds, although it’s length will be different at different locations.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Suggest Corrections