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Andreas93 [3]
3 years ago
11

think about something that has happened to you physically—a fall, a jump, an accident, or something you may have done hundreds o

f times in your favorite sport. Analyze the action and describe it in terms of Newton’s laws. Identify the initial conditions and the forces involved.
Physics
1 answer:
timurjin [86]3 years ago
5 0

My best hobby is driving. Driving has something to do with Newton's first law of motion, which states that an object will continue to be in its state of rest or in uniform motion in a straight line unless it is acted upon by an external force. This law means that an object will continue to be in motion in the same direction unless it is acted upon by a force. Newton's first law of motion is also called the law of inertia.

I usually experience the law of inertia when I am driving my car.

Every morning, for me to move the car from its state of rest to a state of uniform motion, I have to switch on the ignition, which represent an unbalanced force that move the car out of its states of rest. When I am driving, the car continue in motion and in the same direction, unless I apply the brake. The application of the brake is an example of applying an unbalanced force to stop a body in motion.

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For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
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Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

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