Ask & Answer Question

physics ncert solutions class 11th Physics Class 11th Oscillations

14.15 The acceleration due to gravity on the surface of moon is 1.7 m s^–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m s^–2)

3 Views | Posted 5 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

Acceleration due to gravity on Moon surface, g’ = 1.7 m/ $s^{2}$

Acceleration due to gravity on Earth surface, g = 9.8 m/ $s^{2}$

Time period on Earth, T = 3.5 s

We know T = 2 $π \sqrt{\frac{l}{g}}$ where l = length of the pendulum

l = $\frac{T^{2}}{{(2 π}^{2})} \times g$ = $\frac{{3.5}^{2}}{{(2 π}^{2})}$ $\times 9.8$ = 3.041 m

On Moon surface, the length of the pendulum remained same = 3.041 m

So time period on moon surface, T’ = 2 $π \sqrt{\frac{l}{g'}}$ = 2 $π \sqrt{\frac{3.041}{1.7}}$ = 8.40 s

0 Upvotes 0 Downvotes

Similar Questions for you

If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:

Vishal Baghel

$T = 2 π \sqrt[]{\frac{l}{g}}$

$\Rightarrow 2 = 2 π \sqrt[]{\frac{2}{g}}$

$\Rightarrow g = 2 π^{2} m / s^{2}$

In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. New amplitude of oscillation will be:

Vishal Baghel

Velocity of block in equilibrium, in first case,

$v = A ω = A . \sqrt[]{\frac{k}{M}}$

Velocity of block in equilibrium, is second case,

$v' = A' ω' = A' \sqrt[]{\frac{k}{M + m}}$

From conservation of momentum,

Mv = (M + m) v’

$\Rightarrow M A \sqrt[]{\frac{k}{M}} = (M + m) A' \sqrt[]{\frac{k}{M + m}} \Rightarrow A' = A \sqrt[]{\frac{M}{M + m}}$

The fundamental frequency of an organ pipe open at one end is 300 Hz. The frequency of 3rd overtone of n Find n.×this organ pipe is 100 Hz

Vishal Baghel

f? = 300 Hz
3rd overtone = 7f? = 2100 Hz

Consider two identical springs each of spring constant k and negligible mass compared to the mass M as shown. Fig. 1 shows one of them and Fig. 2 shows their series combination. The ratios of time period of oscillation of the two SHM is T_b / T_a = √x, where value of x is ................... (Round off to the Nearest Integer)

Vishal Baghel

Kindly consider the following figure

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

Vishal Baghel

K = U

½ mω² (A² - x²) = ½ mω²x²

A² - x² = x²

A² = 2x²

x = ± A/√2

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
686k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

14.15 The acceleration due to gravity on the surface of moon is 1.7 m s^–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m s^–2)

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

14.15 The acceleration due to gravity on the surface of moon is 1.7 m s–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m s–2)

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

14.15 The acceleration due to gravity on the surface of moon is 1.7 m s^–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m s^–2)