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zimovet [89]
3 years ago
7

The mass of a mass-and-spring system is displaced 10 cm from its equilibrium position and released. A frequency of 4.0 Hz is obs

erved. What frequency would be observed if the mass had been displaced only 5.0 cm and then released?
Physics
2 answers:
scoundrel [369]3 years ago
7 0

Answer:

It will remain same i.e. 4.0 Hz

Explanation:

In a mass-and-spring system, the frequency depends upon the mass (m) and the spring constant (k).

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

Since, the frequency does not depend on the initial displacement of the mass, the frequency would remain the same i.e. 4.0 Hz.

Tanya [424]3 years ago
7 0

Answer:

f = 4.0 Hz

Explanation:

The frequency of SHM in spring mass system is given as

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

now we can see that the frequency of SHM of spring mass system depends upon following parameters

1) mass of the object

2) spring constant

So here it does not depends on the displacement of the object from its mean position

So here the frequency will remain the same

So here

f = 4.0 Hz

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Let car A's starting position be the origin, so that its position at time <em>t</em> is

A: <em>x</em> = (40 m/s) <em>t</em>

and car B has position at time <em>t</em> of

B: <em>x</em> = 100 m - (60 m/s) <em>t</em>

<em />

They meet when their positions are equal:

(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>

(100 m/s) <em>t</em> = 100 m

<em>t</em> = (100 m) / (100 m/s) = 1 s

so the cars meet 1 second after they start moving.

They are 100 m apart when the difference in their positions is equal to 100 m:

(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m

(subtract car B's position from car A's position because we take car A's direction to be positive)

(100 m/s) <em>t</em> = 200 m

<em>t</em> = (200 m) / (100 m/s) = 2 s

so the cars are 100 m apart after 2 seconds.

3 0
2 years ago
How far does a runner run if he runs for 60 seconds at 5 m/s
mylen [45]

Answer: 20 miles

Explanation:

4 0
2 years ago
Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(
bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude  = 2.20 mm

b) frequency =

     f = \dfrac{\omega}{2\pi}

     f = \dfrac{743}{2\pi}

          f = 118.25 Hz

c) wavelength

        k= \dfrac{2\pi}{\lambda}

        \lambda= \dfrac{2\pi}{k}

        \lambda= \dfrac{2\pi}{7.02}

        \lambda= 0.895\ m

d) speed

         v = \dfrac{\omega}{k}

         v = \dfrac{743}{7.02}

                v = 105.84 m/s

e) direction of the motion will be in negative x-direction

f) tension

  T = \dfrac{v^2\ m}{L}

  T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}

      T = 27.87 N

g) Power transmitted by the wave

  P = \dfrac{1}{2}m\ v \omega^2\ A^2

  P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2

      P = 0.438 W

5 0
2 years ago
A 25 kg rock resting on the bottom of a lake must be moved from the paths of boats. The rock has a density of 2350 kg/m^3. What
algol13

Answer:

The force needed is the weight of the rock minus the buoyant force.

Explanation:

7 0
2 years ago
002 10.0 points
snow_lady [41]

Answer:

-2040 m/s²

Explanation:

Taking toward the wall to be positive, the initial velocity is 10.1 m/s and the final velocity is -8.3426 m/s.

Average acceleration is the change in velocity over change in time.

a = Δv / Δt

a = (-8.3426 m/s − 10.1 m/s) / 0.00905 s

a = -2040 m/s²

3 0
2 years ago
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