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iren [92.7K]
2 years ago
12

A damped harmonic oscillator consists of a block of mass 2.5 kg attached to a spring with spring constant 10 N/m to which is app

lied a damping force (in Newtons) of the form F = –0.1v, with v the velocity in m/s. The spring is stretched a distance xm and released. After four complete oscillations, what fraction of the mechanical energy is retained by the system?

Physics
1 answer:
Cerrena [4.2K]2 years ago
6 0

Answer:

0.5% per oscillation

Explanation:

The term 'damped oscillation' means an oscillation that fades away with time. For Example; a swinging pendulum.

Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).

Where m= Mass, v= velocity.

Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).

Where k= force constant, X= displacement.

Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).

Therefore, we use equation (3) to get the resonance frequency,

W^2= k/m--------------------------------------------------------------------------------------(3)

Slotting values into equation (3).

= 10/2.5.

= ✓4.

= 2 s^-1.

Recall that, F= -kX

F^2= (-0.1)^2

Potential energy,PE= 1/2 ×0.01

Potential energy= 0.05 ×100

= 0.5% per oscillation.

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An older camera has a lens with a focal length of 60mm and uses 34-mm-wide film to record its images. Using this camera, a photo
lesya692 [45]

Answer:

24.71 mm

Explanation:

Distance is proportional to focal length, so

d∝f

which means

\frac{d'_1}{d'_2}=\frac{f_1}{f_2}

Magnification of first lens

M_2=-\frac{d'_1}{d_1}

                   and

M_2=\frac{h'_1}{h_1}

Similarly, magnification of second lens

M_2=-\frac{d'_2}{d_1}

                   and

M_2=\frac{h'_2}{h_1}

From the above equations we get

\frac{M_1}{M_2}=\frac{d'_1}{d_2'}

                   and

\frac{M_1}{M_2}=\frac{h'_1}{h_2'}

which means,

\frac{d'_1}{d_2'}=\frac{h'_1}{h_2'}

and

\frac{d'_1}{d_2'}=\frac{f_1}{f_2}

So, we get

\frac{f_1}{f_2}=\frac{h'_1}{h_2'}\\\Rightarrow f_2=f_1\times\frac{h_2'}{h'_1}\\\Rightarrow f_2=60\times\frac{14}{34}=24.71\ mm

∴ Focal length should this camera's lens is 24.71 mm

6 0
2 years ago
Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic
Nuetrik [128]

Answer:

(a) A = 3.90 \AA

(b) A = 4.50 \AA

(c) A = 5.51 \AA

(d) A = 9.02 \AA

Solution:

As per the question:

Radius of atom, r = 1.95 \AA = 1.95\times 10^{- 10} m

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = 2\times 1.95 = 3.90 \AA

(b) For body centered cubic lattice:

A = \frac{4}{\sqrt{3}}r

A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA

(c) For face centered cubic lattice:

A = 2{\sqrt{2}}r

A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA

(d) For diamond lattice:

A = 2\times \frac{4}{\sqrt{3}}r

A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA

6 0
2 years ago
A hot air balloon of total mass M (including passengers and luggage) is moving with a downward acceleration of magnitude a. As i
LUCKY_DIMON [66]

Answer:

The fraction of mass that was thrown out is calculated by the following Formula:

M - m = (3a/2)/(g²- (a²/2) - (ag/2))

Explanation:

We know that Force on a moving object is equal to the product of its mass and acceleration given as:

F = ma

And there is gravitational force always acting on an object in the downward direction which is equal to g = 9.8 ms⁻²

Here as a convention we will use positive sign with acceleration to represent downward acceleration and negative sign with acceleration represent upward acceleration.

Case 1:

Hot balloon of mass = M

acceleration = a

Upward force due to hot air = F = constant

Gravitational force downwards = Mg

Net force on balloon is given as:

Ma = Gravitational force - Upward Force                              

Ma = Mg - F                      (balloon is moving downwards so Mg > F)

F = Mg - Ma

F = M (g-a)

M = F/(g-a)

Case 2:

After the ballast has thrown out,the new mass is m. The new acceleration is -a/2 in the upward direction:

Net Force is given as:

-m(a/2) = mg - F        (Balloon is moving upwards so F > mg)

F = mg + m(a/2)

F = m(g + (a/2))

m = F/(g + (a/2))

Calculating the fraction of the initial mass dropped:

M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]

5 0
2 years ago
The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17°C and
mina [271]

Answer:

\frac{dQ}{dt}= 4312 W

Explanation:

As we know that base of the slab is given as

A = 11 \times 8

A = 88 m^2

now we know that rate of heat transfer is given as

\frac{dQ}{dt} = \frac{kA}{x} (T_2 - T_1)

here we know that

k = 1.4 W/m k

Also we have

x =0.20

\frac{dQ}{dt} = \frac{1.4(88)}{0.20}(17 - 10)

\frac{dQ}{dt}= 4312 W

7 0
2 years ago
A nonlinear spring is used to launch a toy car. The car is pushed against the spring, compressing the spring 2.5 cm. The force t
dybincka [34]

Answer:

1.56 J

Explanation:

given,

Spring compression, x = 2.5 cm

Force exerts by the spring,

F = - k x

k = 5000 N/m

Potential energy stored = ?

energy stored in the spring

PE = \dfrac{1}{2}kx^2

PE = \dfrac{1}{2}\times 5000\times 0.025^2

PE = 1.56 J

Hence, the potential energy stored in the car is equal to 1.56 J.

5 0
2 years ago
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