All categories

3 years ago

If the spring constant is doubled, what value does the period have for a mass on a spring?

Physics

2 answers:

quester [9]3 years ago

5 0

Answer:

The period would decrease by `sqrt(2)`.

Explanation:

The angular frequency omega of an oscillating mass m due to a spring with a constant k is given by

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

(this is obtained by solving the differential equation $m\ddot{{ x}}-kx=0$ )

If k doubles, i.e., k'=2k, then

$\omega'=\sqrt{\frac{k'}{m}}=\sqrt{\frac{2k}{m}}=\sqrt{2}\sqrt{\frac{k}{m}}=\omega\sqrt{2}$

Since the angular frequency is $\omega = \frac{2\pi}{T}$ , we can say that

$\omega\sqrt{2}=\frac{2\pi}{T}\sqrt{2}=\frac{2\pi}{\frac{T}{\sqrt{2}}}=\frac{2\pi}{T'}$

and so it becomes clear that the period T will decrease by sqrt(2) as stated in choice (D).

lisabon 2012 [21]3 years ago

5 0

Answer:

D. The period would decrease by `sqrt(2)`. is wrong on clever/plato

Explanation:

i got it wrong :(

You might be interested in

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

7 0

3 years ago

A ball of mass m= 450.0 g traveling at a speed of 8.00 m/s impacts a vertical wall at an angle of θi =45.00 below the horizontal

Mkey [24]

Answer:

F = 20.4 i ^

Explanation:

This exercise can be solved using the ratio of momentum and amount of movement.

I = F t = Dp

Since force and amount of movement are vector quantities, each axis must be worked separately.

X axis

Let's look for speed

cos 45 = vₓ / v

vₓ = v cos 45

vₓ = 8 cos 45

vₓ = 5,657 m / s

We write the moment

Before the crash p₀ = m vₓ

After the shock $p_{f}$ = -m vₓ

The variation of the moment Δp = mvₓ - (-mvₓ) = 2 m vₓ

The impulse on the x axis Fₓ t = Δp

Fₓ = 2 m vₓ / t

Fx = 2 0.450 5.657 / 0.250

Fx = 20.4 N

We perform the same calculation on the y axis

sin 45 = vy / v

vy = v sin 45

vy = 8 sin 45

vy = 5,657 m / s

We calculate the initial momentum po = m $v_{y}$

Final moment $p_{f}$ = m $v_{y}$

Variations moment Δp = m $v_{y}$ - m $v_{y}$ = 0

Force in the Y-axis $F_{y}$ = 0

Therefore the total force is

F = fx i ^ + Fyj ^

F = Fx i ^

F = 20.4 i ^

3 0

4 years ago

A uniform rod is attached to a wall by a hinge at its base. The rod has a mass of 6.5 kg, a length of 1.3 m, is at an angle of 1

liubo4ka [24]

Answer:

tension wire 104 N, horizontal force hinge 104 N, vertical force hinge 63.7 N

Explanation:

The system described is in static equilibrium under the action of several forces, which are shown in the attached diagram, where T is the tension of the wire, W is the weight of the bar, Fx is the horizontal reaction of the wall and Fy It is the vertical reaction of the wall.

The directions of the forces are indicated by the arrows and are marked intuitively, but when solving the problem if one gives a negative value this indicates that the direction is wrong, but does not alter the results of the problem

For the resolution we use Newton's second law, both in translational and rotational equilibrium, if necessary.

We must establish a reference system to assign the positive meaning, we place it with the origin in the hinge, and the positive directions to the right and up. The location of the coordinate system allows us to eliminate the reaction of the hinge by having zero distance to origin. We write the equilibrium equations for each axis

∑ Fx = 0

Fx -T =0

∑ Fy =0

Fy -W =O W = m g

∑τ =0

-T dy + W dx =0

For rotational equilibrium we take the positive direction as counterclockwise rotation and distance is the perpendicular distance of the force to the axis of the coordinate system

dy = L sin 17

dx = (L/2) cos 17

Where L is the length of the bar and the weight is applied at the center of it, we write and simplify the equation

-T L sin 17 + mg (L / 2) cos 17 = 0

- T sin 17 + mg/2 Cos 17 =0

We write the three equations together

Fx- T = 0

Fy -W = 0

-T sin 17 + (m g / 2) cos 17 = 0

a) With the third equation we can find the wire tension

T = m g / 2 cos 17 / sin17

T = 6.5 9.8/2 cotan 17

T = 104 N

b) We use the first equation to find Fx

Fx- T =0

Fx = T = 104 N

c) We use the second equation to find Fy

Fy = W = m g

Fy = 6.5 9.8

Fy = 63.7 N

4 0

3 years ago

A car is left overnight in a parking lot. What is its instantaneous speed at 12:01 a.m.?

STatiana [176]

Answer:

0 mph

Explanation:

4 0

3 years ago

Read 2 more answers

A change in temperature causes a change in density. true or false?

zvonat [6]

True a change in temperature can cause a change in density.

6 0

3 years ago

Read 2 more answers