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Ratling [72]
3 years ago
6

A 16 pound weight attached to a spring exhibits simple harmonic motion. Determine the equation of motion if the spring constant

is 8 lb/ft and if the weight is released 6 inches below the equilibrium position with a downward velocity of 1 ft/s.
Physics
1 answer:
yarga [219]3 years ago
8 0

Answer:

T = 2.82π s

x = Acos(0.71t)

Explanation:

This problem can be solved by using the expressions

T = 2\pi \sqrt{\frac{m}{k} }    ( 1 )

x=Acos(\omega t) = Acos(\frac{2\pi }{T}t )   ( 2 )

where T is the period of oscillation of the system, m is the mass of the object attached to the spring, k is the spring constant and x is the position of the object.

By replacing in the expression (1):

T=2\pi \sqrt{\frac{16 lb}{8lb/ft}} = 2.82\pi s

Taking 6 inches as the amplitude of the motion, we have

x=6cos(\frac{2\pi }{2.82\pi }t ) = 6cos(0.71t)

I hope this is useful for you

Regards

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A jogger runs at a constant rate of 10.0 m every 2.0 seconds. The jogger starts at the origin and runs in the positive direction
Elis [28]

Answer:

(a) 25 m

(b) 75 m

Explanation:

Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.

So, the speed of the jogger,

v=\frac{10}{2}=5m/s\;\cdots(i)

Let d be the distance covered by him in time, t s.

As distance=(speed) x (time)

So, d=vt

From equation (i)

\Rightarrow d=5t\;\cdots(ii)

As the jogger starts from origin, so, the distance, d, also represents the position of the jogger at the time t s.

The position-time graph has been shown.

(a) From equation (ii), for t=5.0 s

d=5\times 5=25 m

So, the jogger is at a distance of 25 m from the origin.

(b) Similarly, for t=15.0 s

d=5\times 15=75 m

So, the jogger is at a distance of 75 m from the origin.

8 0
3 years ago
A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo
m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is   spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 .  t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

    p₀ = m1 v₀₁

Moment after shock

    p_{f} = (m1 + m2) v_{f}

   p₀ = p_{f}

   m1 v₀₁ = (m1 + m2) v_{f}

  v_{f} = v₀₁ m1 / (m1 + m2)

   v_{f}= 4.4 600 / (600 + 500)

  v_{f} = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

   Em₀ = K = ½ (m1 + m2) v_{f}²

After compressing the spring

   E_{mf} = Ke = ½ k x²

As there is no rubbing the energy is conserved

   Em₀ = E_{mf}

   ½ (m1 + m2) v_{f}² = = ½ k x²

   x = v_{f} √ (k / (m1 + m2))

   x = 2.4 √ (11/3000)

   x = 0.396 m

7 0
3 years ago
You are creating waves in a rope by shaking your hand back and forth. Without changing the distance your hand moves, you begin t
scoundrel [369]

Answer:

<u>Amplitude - remains the same</u>

<u>Frequency - increases</u>

<u>Period - decreases</u>

<u>Velocity - remains the same.</u>

<u />

Explanation:

The amplitude of the wave remains the same since you are not changing the distance your hand moves and the amplitude of the wave depends on how much distance your hand covers while moving.

The frequency of your wave increases since now you are moving your hand more number of times in the same period i.e. your hand is moving faster in one second. So, the frequency of your wave increases.

The period is the time taken by the wave to travel a certain distance. Since your hand is now moving faster, the wave will travel faster and will take less time to cover the same distance hence, we can say that its period will decrease.

The velocity of a wave depends on the medium in which it is travelling. Your wave was previously travelling in air and the new wave is also travelling in the same medium so the velocity of the wave remains unchanged.

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Answer:

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2 years ago
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Electricity. I took something like this hope this helps :)
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