Question

A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T. If the bob's mass is doubled, approximately what will the pendulum's new period be? A: T/2 B: T C: sqrt(2)*T D: 2T ---------------- If the pendulum is brought on the moon where the gravitational acceleration is about g/6, approximately what will its period now be? A: T/6 B: T/sqrt(6) C: sqrt(6)*T D: 6T ---------------- If the pendulum is taken into the orbiting space station what will happen to the bob? A: It will continue to oscillate in a vertical plane with the same period. B: It will no longer oscillate because there is no gravity in space. C: It will no longer oscillate because both the pendulum and the point to which it is attached are in free fall. D: It will oscillate much faster with a period that approaches zero

Answer 1

The right question is

b.) If the pendulum is carried to the moon where the acceleration of gravity is around g / 6, what is the current period?

Further Explanation

A simple pendulum consists of a light string and a small ball (pendulum ball) with mass m hanging from the end of the rope. In analyzing the movement of a simple pendulum, the air friction force is ignored and the mass of the rope is so small that it can be ignored relative to the ball.

A simple pendulum consisting of a rope with a length L and a pendulum ball with mass m. The forces acting on the pendulum ball are the weight force (w = mg) and the FT string tension force. Gravity has a component of mg cos theta which is in the direction of the rope and mg sin theta which is perpendicular to the rope. The pendulum oscillates due to the presence of mg sin theta gravity component. Because there is no air friction, the pendulum oscillates along a circular arc with the same amplitude.

The requirement for an object to do Simple Harmonic Motion is if the recovery force is proportional to the deviation. If the recovery force is proportional to the deviation of x or the angle of the theta, the pendulum performs Simple Harmonic Motion.

The simple pendulum period can be determined using the equation:

T = 2n (sqrt m / k

We replace the effective force constant with mg / L

T = 2n (sqrt m / (mg / L))

T = 2n (sqrt L / g -> 0 small)

Simple Pendulum Frequency

f = 1 / T

f = 1 / 2n (sqrt L / g)

f = (1 / 2n) (sqrt g / L -> 0 small)

This is a simple pendulum frequency equation

Information :

T is the period, f is the frequency, L is the length of the rope and g is the acceleration due to gravity.  

Learn More

Simple Pendulum  brainly.com/question/12473773

Formula simple pendulum brainly.com/question/12644845

Detail

Class: High School

Subject: Physics

Keywords: pendulum, simple, formula

Answer 2

1. B. T

The period of a simple pendulum is given by:

$T=2\pi \sqrt{\frac{L}{g}}$ (1)

where

L is the length of the pendulum

g is the gravitational acceleration

From the formula, we notice that the period of the pendulum does not depend on the mass of the bob. Therefore, when the bob's mass is doubled, the period does not change.

2. C: sqrt(6)*T

In this case, the pendulum is brought to the moon, where the gravitational acceleration is

$g'=\frac{g}{6}$

If we substitute this value into the equation for the period (1), we find the new period of the pendulum:

$T'=2\pi \sqrt{\frac{L}{g'}}=2\pi \sqrt{\frac{L}{(g/6)}}=\sqrt{6}(2\pi \sqrt{\frac{L}{g}})=\sqrt{6} T$

3. B: It will no longer oscillate because there is no gravity in space

Explanation:

The motion (oscillation) of the pendulum is caused by the force of gravity, which "pulls" the bob towards the equilibrium position. If there is no gravity, then there is no force acting on the bob, therefore the pendulum can no longer oscillate.

So, the correct answer is

B: It will no longer oscillate because there is no gravity in space