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3 years ago

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A horizontal spring-mass system has low friction, spring stiffness 205 N/m, and mass 0.6 kg. The system is released with an init

ial compression of the spring of 13 cm and an initial speed of the mass of 3 m/s.
(a) What is the maximum stretch during the motion? m
(b) What is the maximum speed during the motion? m/s
(c) Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system. What is the average power input in watts required to maintain a steady oscillation?

1 answer:

Andreyy893 years ago

5 0

Answer:

(a). The maximum stretch during the motion is 20.7 cm

(b). The maximum speed during the motion is 3.84 m/s.

(c). The energy is 0.060 Watt.

Explanation:

Given that,

Spring stiffness = 205 N/m

Mass = 0.6 kg

Compression of spring = 13 cm

Initial speed = 3 m/s

(a). We need to calculate the maximum stretch during the motion

Using conservation of energy

$E_{initial}=E_{final}$

$\dfrac{1}{2}kx_{c}^2+\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_{m}^2$

Put the value into the formula

$\dfrac{1}{2}\times205\times(13\times10^{-2})^2+\dfrac{1}{2}\times0.6\times3^2=\dfrac{1}{2}\times205\times x_{m}^2$

$x_{m}=\sqrt{\dfrac{4.43\times2}{205}}$

$x_{m}=20.7\ cm$

(b). Maximum speed comes when stretch is zero.

We need to calculate the maximum speed during the motion

Using conservation of energy

$E_{initial}=E_{final}$

$\dfrac{1}{2}kx_{c}^2+\dfrac{1}{2}mv^2=\dfrac{1}{2}mv'^2$

Put the value into the formula

$\dfrac{1}{2}\times205\times(13\times10^{-2})^2+\dfrac{1}{2}\times0.6\times3^2=\dfrac{1}{2}\times0.6\times v'^2$

$v'=\sqrt{\dfrac{4.43\times2}{0.6}}$

$v'=3.84\ m/s$

(c). Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system

We need to calculate the time period

Using formula of time period

$T=2\pi\sqrt{\dfrac{m}{k}}$

Put the value into the formula

$T=2\pi\sqrt{\dfrac{0.6}{205}}$

$T=0.33\ sec$

We need to calculate the energy

Using formula of energy

$E=\dfrac{P}{t}$

Put the value into the formula

$E=\dfrac{0.02}{0.33}$

$E=0.060\ Watt$

Hence, (a). The maximum stretch during the motion is 20.7 cm

(b). The maximum speed during the motion is 3.84 m/s.

(c). The energy is 0.060 Watt.

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1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

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$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

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