Question

2. [10 points] A block of mass 0.5 kg is attached to a spring and is moving in simple harmonic motion on a frictionless horizontal surface. The amplitude of the motion is 0.25 m and the period is 2.0 s. When the block is at x = 0.1 m, what are (a) the speed, (b) the acceleration, (c) the potential energy and (d) the kinetic energy of the block? And what are (e) the total energy, (f) the maximum speed and (g) the maximum acceleration of the block?

          2. [10 points] A block of mass 0.5 kg is attached to a spring and is moving in simple harmonic motion on a frictionless horizontal surface. The amplitude of the motion is 0.25 m and the period is 2.0 s. When the block is at x = 0.1 m, what are
(a) the speed,
(b) the acceleration,
(c) the potential energy and
(d) the kinetic energy
of the block? And what are
(e) the total energy,
(f) the maximum speed and
(g) the maximum acceleration of the block?

Show more…

2. [10 points] A block of mass 0.5 kg is attached to a spring and is moving in simple harmonic motion on a frictionless horizontal surface. The amplitude of the motion is 0.25 m and the period is 2.0 s. When the block is at x = 0.1 m, what are
(a) the speed,
(b) the acceleration,
(c) the potential energy and
(d) the kinetic energy
of the block? And what are
(e) the total energy,
(f) the maximum speed and
(g) the maximum acceleration of the block?

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University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Pritesh Ranjan

11/15/2021

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Transcript

00:01 In this problem, we have been given that there is a block and this block is having mass of 0 .5 kilograms.

00:07 So this block is connected to the end of a spring that is oscillating.

00:13 So the block is oscillating here such that it's kept on a horizontal surface which is smooth.

00:20 And it is observed that the amplitude of oscillation is 0 .25 meters.

00:26 Also, we have been given that the time period of oscillation is two seconds.

00:32 So let's consider k as the spring factor.

00:35 And when the block is present at 0 .1 meters, at this point we have to find out various things.

00:44 So first, we can observe that omega is equal to root of k by m.

00:53 And omega, we know it is 2 pi by t.

00:55 So the time period will be 2 pi root m by k.

00:59 So let's figure out the spring factor first.

01:02 So putting the values that we have been given, 2 will be equal to 2 pi root of 0 .5 by k.

01:08 So from here we can square both sides to get the value of k.

01:11 So that will come out to be 4 pi square into 0 .5 divided by 4.

01:17 And that gives us the value of k as pi square into 0 .5.

01:21 So approximately pi square is 10.

01:23 So we take the value of k as 5 newtons per meter.

01:28 And when the block is present at point 1 meter, here we need to determine the speed.

01:36 So the expression that relates the speed is equal to omega root of a square minus x square.

01:45 So omega here can be computed because we have been given the time period.

01:49 So let's compute the omega so that will be 2 pi by t.

01:52 So this comes out to be pi radiance per second...

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