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Alex777 [14]
4 years ago
11

A simple undamped spring-mass system is set into motion from rest by giving it an initial velocity of 100 mm/s. It oscillates wi

th a maximum amplitude of 10 mm. What is its natural frequency?
Engineering
1 answer:
____ [38]4 years ago
6 0

Answer:

f=1.59 Hz

Explanation:

Given that

Simple undamped system means ,system does not consists any damper.If system consists damper then it is damped spring mass system.

Velocity = 100 mm/s

Maximum amplitude = 10 mm

We know that for a simple undamped system spring mass system

V_{max}=\omega A

now by putting the values

V_{max}=\omega A

100 = ω x 10

ω = 10 rad/s

We also know that

ω=2π f

10 = 2 x π x f

f=1.59 Hz

So the natural frequency will be f=1.59 Hz.

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Answer:

True

Explanation:

When trying to solve a frame problem in Engineering or Physics, it will typically be necessary to draw more than one body diagram.

When we have several parts of the frame or a set of frames, we have the anchor point, as well as the intersections of frames. Besides that, usually, there is a particle or rigid body together with the frame system. In this sense, usually, it is required to analyze a body diagram for the particle or rigid body suspended, as well as the intersections of the frames. So, usually, it will be required a minimum of two body diagrams.

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3 years ago
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Answer:

a) t = 277.477\,s\,(4.625\min), b) v_{f} = 0\,\frac{mi}{h}, c) a = -0.128\,\frac{ft}{s^{2}}

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b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be v_{f} = 0\,\frac{mi}{h}.

c) The final constant deceleration is:

a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}

a = -0.128\,\frac{ft}{s^{2}}

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