Physics, 06.05.2020 02:10 july00

A simple pendulum and a mass hanging on a spring both have a period of 1 sec when set into oscillatory motion on Earth. They are taken to Planet X, which has the same diameter as Earth but twice the mass. Which is true about the periods of the two objects on Planet X compared to their periods on Earth?

a. Both are shorter. Both are the same.
b. The period of the spring is shorter;
c. the pendulum is the same.
d. The period of the pendulum is shorter;
e. the spring is the same

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Answer from: brittanysanders

d.The period of the pendulum is shorter;

e.the spring is the same

The period of a simple pendulum is given by:

$T=2\pi\sqrt{\frac{l}{g}}$

Where l is the pendulum length and g is the gravitational aceleration of the planet.

The period of a mass-spring system is given by:

$T=2\pi\sqrt\frac{m}{k}$

Where k is the spring constant.

The gravitational acceleration on the Earth's surface is defined as:

$g=G\frac{M}{r^2}$

Here G is the Cavendish constant, M is the Earth mass and r its radius.

For Planet X, we have r'=r and M'=2M . Thus, the gravity acceleration on this planet is:

$g'=G\frac{M'}{r'^2}\\g'=G\frac{2M}{r^2}\\g'=2(G\frac{M}{r^2})\\g'=2g$

Since the period of the pendulum is inversely proportional to gravity, the period of the pendulum is shorter on Planet x. In other hand, the period of the mass-spring system is the same in both planets.

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

Answer from: Quest

The answer is 4f======since electrostatic force is given by $f_{elec} = \dfrac{kq_1 q_2}{r^2}$ we let f be the electrostatic force when r = 1 m $f = \dfrac{kq_1 q_2}{1^2} = kq_1q_2$ if we let $f_{0.5}$ denote the electrostatic force when the distance is decreased so that r = 0.5m, then $\begin{aligned} f_{0.5} & = \dfrac{kq_1 q_2}{0.5^2} \\ & = \frac{1}{0.5^2} \cdot kq_1q_2 \\ & = \frac{1}{\left( \frac{1}{2}\right)^2} \cdot kq_1q_2 \\ & = 4 \cdot \boxed{ kq_1q_2 } \end{aligned}$ we established earlier that f = kq₁q₂. therefore $f_{0.5} = 4 f$ the force will now be 4f.

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