Physics, 09.11.2019 07:31 Queenofpizza

Ablock attached at the end of a spring undergoes simple harmonic motion with frequency of oscillation ω.

which of the following modifications will make the block oscillate with a greater frequency?

(a) increase the amplitude of the simple harmonic motion
(b) increase the spring constant
(c) increase the mass of the block
(d) increase the phase angle

Answers: 2

Show answers

Answers

Answer from: reginaldboyd28

Explanation:

Let $\omega$ is the frequency of oscillation of a block that undergoes simple harmonic motion. The frequency of oscillation is given by :

$\omega=\sqrt{\dfrac{k}{m}}$ ...............(1)

Where

k is the spring constant of the spring

m is the mass of the block

It is clear from equation (1) that the frequency of the oscillation depends on the spring constant and the mass of the block. So, on increasing the spring constant, the block oscillate with a greater frequency.

Answer from: nikkio4

(B) Increase the spring constant

Explanation:

The frequency os oscillation of a block is given by the following formula:

$w = \sqrt{\frac{k}{m}}$

In which k is the spring constant and m is the mass of the block.

So the answer is either b or c, as the frequency of oscillation is a function of the sping constant and of the mass of the block.

Lets try b first, increasing the spring constant.

For example, with m = 4.

If k = 16, w = 2.

If k = 36, w = 3

So as the sping constant increases, so does the frequency of oscillation.

So the correct answer is:

(B) Increase the spring constant

Why c is wrong?

Suppose we have k = 36.

if m = 4, w = 3

if m = 9, w = 2

So, as the mass increases, the frequency decreases. So c is wrong

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

Answer from: Quest

Sorry i don’t know what to do

Another question on Physics

Physics, 22.06.2019 07:00

How do you tell if someone actually cares about you? and just say they do?

Answers: 2

Answer

Physics, 22.06.2019 10:00

One object has a mass of 1 kg and another object has a mass of 3 kg. if the speeds are the same, which of the following is true about their kinetic energy?

Answers: 2

Answer

Physics, 22.06.2019 14:00

Amass attached to a spring is displaced from its equilibrium position by 5cm and released. the system then oscillates in simple harmonic motion with a period of 1s. if that same mass–spring system is displaced from equilibrium by 10cm instead, what will its period be in this case? a mass attached to a spring is displaced from its equilibrium position by and released. the system then oscillates in simple harmonic motion with a period of . if that same mass–spring system is displaced from equilibrium by instead, what will its period be in this case? a) 0.5sb) 2sc) 1sd) 1.4s

Answers: 2

Answer

Physics, 22.06.2019 22:50

Cyclotrons are widely used in nuclear medicine for producing short-lived radioactive isotopes. these cyclotrons typically accelerate h- (the hydride ion, which has one proton and two electrons) to an energy of 5 mev to 20 mev. this ion has a mass very close to that of a proton because the electron mass is negligible—about 1 2000 of the proton’s mass. a typical magnetic field in such cyclotrons is 1.9 t. (a) what is the speed of a 5.0-mev h-? (b) if the h- has energy 5.0 mev and b = 1.9 t, what is the radius of this ion’s circular orbit?

Answers: 1

Answer

You know the right answer?

Ablock attached at the end of a spring undergoes simple harmonic motion with frequency of oscillatio...

Questions