T = 2pi * √L/g
on that planet
T' = 2pi √( L /g/6)
= √(6L/g)
= √6 * √L/g
=√6*T A option
T-20 e 11. On a planet A, the acceleration of gravity is g/6. If a pendulum...
What is the acceleration due to gravity on a planet where a 1.5 m long pendulum oscillates with the same period as a 0.8 kg mass vibrates on an 80 N/m spring?
A simple pendulum on the surface of the Earth has period T. On a different planet, the same pendulum has period 27. The free-fall acceleration on this planet is 04g O g/4 O2g O g/2
Part 1: Pendulum on allen planet You are on a mysterious planet and notice that a 2.7 m long simple pendulum has a period of 3s when its angular amplitude is " or less. A) Determine the value for the free fall acceleration on the surface of this planet. B.) If you DOUBLED the length of the pendulum, what would its period be? Part 2: Pendulum on Earth You are on Earth and notice that the pendulum in grandfather clock...
1. How much is the period of 1=1.00 m long pendulum on the Moon (g = 1.600 m/s2) 4.97 sec. 2. On a planet X pendulum of the length 0.500 m makes 50.0 oscillations in 1.00 min. Find the acceleration of gravity on the planet X. g = 13.7 m/s2. 3. Find the period of small oscillations of a meter stick suspended by its end near Earth surface (assume g=9.800 m/s2). Notice that this is not a simple pendulum but...
o) The acceleration due to gravity on the moon is 1/6 of what it is on Earth. We are going to do clock experiments on the moon. We have one clock that runs because of the oscillation of a ass (m) on an ideal spring (spring constant k), which oscillates with an amplitude A, We have a second clock that runs because of a simple pendulum made of a mass (m) on a wire of length L, which oscillates with...
28. In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to (a) 0.7% (b) 6.8%...
Consider the pendulum shown at the right. It has the following characteristics: • T = 1.33 s • m=0.251 kg Answer the three questions below, using three significant digits. Part A: What is the value of the angular frequency (w) of the pendulum? T. 4.72 rad/s Part B: What is the length (L) of the pendulum? L = .44 m Part C: How might the period of this pendulum be different if it were moved from Earth to the planet...
The acceleration of gravity is 9.81 m/s How 'l is the tower? Answer in units of m. 004 10.0 points A mass of 0.22 kg is attached to a spring and is set into vibration with a period of 0.22 s. What is the spring constant of the spring? Answer in units of N/m 005 (part 1 of 6) 10.0 points Consider a pendulum of length 4.488 m. The acceleration of gravity is 9.832 m/s a) What is its period...
(a) What is the length of a simple pendulum that oscillates with a period of 3.0 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2? LE = Lм %3 m (b) What mass would you need to suspend from a spring with a force constant of 20 N/m in order for the mass-spring system to oscillate with a period of 3.0 s on Earth, where...
Calculus question please help
<3 . (ignore the working)
4. The period of a pendulum is given by T = 2 π l-where l is the length of the pendulum and g is the acceleration due to gravity. Suppose I = 5 feet feet with a maximum error of 0.01 feet .01 feet and T = 2 seconds with a maximum error of 0. 05 seconds Use differentials to estimate the maximum error of g Hint: solve for g first....