Question

A point body with mass 2.17 kg hangs in a massless string of length 3.14m. We pull the point body a little angle to the side and release. At what frequency will the point body commute?

In this assignment we count without a name. The position of a body is given by:
x (t) = 7.210 + 8,031t-4,905t ^ 2
Will the body ever have zero acceleration? If so, when?

A bullet is ejected from the ground with a starting speed of 27m / s and an angle of 72º to the horizontal plane. After how long will the ball be 5m above the ground?

Answer 1

The above system can be called as a pendulum.

The equation of the frequency of a pendulum=

f= 1/2π *√(g/l)

Where g is acceleration due to gravity and l is the length of the string.

So frequency of oscillation f=

1/2π * √ (9.8/ 3.14)= 0.281 Hz.

The mass will have a zero acceleration.

At the equilibrium point ( centre point) the mass would have maximum velocity but have a zero acceleration.

When the ball is released from the ground, it has 2 velocities, v sin a along vertical direction and . v cos a along horizontal direction.

So the initial vertical velocity, u= v sin a

= 27 sin 72 = 25.67m/s

We have vertical displacement, s= 5 metres

We have the equation of motion

S= u* t + a t²/2, where a is acceleration and t is time taken.

Subtituting the values,

5= 25.67 * t + -9.8* t²

So solving the equation,

We get time , t= 0.211 sec and t= 2.40 sec. Which are the 2 time instance where ball passed through 5 m vertical distance ( on ascend and descend)

So on the ascending motion, the time taken by the ball to reach 5 m above ground = 0.211 sec.