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Alex73 [517]
3 years ago
14

If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. The

object is displaced a distance 0.120 m from its equilibrium position and released with zero initial speed. Then after a time 0.800 s, its displacement is found to be a distance 0.120 m on the opposite side, and it has passed the equilibrium position once during this interval. a) Find the amplitude. b) Find the period. c) Find the frequency."
Physics
2 answers:
goldfiish [28.3K]3 years ago
6 0

Answer:

a)0.120m

b)1.60s

c)0.625Hz

Explanation:

a)In this problem, initial displacement A=0.120m was displaced and released with zero initial speed.

The amplitude is the maximum displacement from equilibrium and it was maximum at the moment t=0.(∴The object goes from x=+A to x=-A and back to complete one cycle)

So the amplitude will be equal to the initial displacement A=0.120m

Therefore, the amplitude is 0.120m

b) movement from +A( max positive displacement) to -A (max negative displacement) took place during half the period of SHM

therefore,

0.800 s= T/2   ---> by further solving for T

T= 0.800 x 2

T=1.6 s

The period is 1.6 s

c) As we know that frequency is inversely proportional to the time period.

To find frequency:

f=1/T           (∴T=1.6)

f=1/1.6

f=0.625Hz

therefore, the frequency is 0.625 Hz

Natali [406]3 years ago
6 0

Answer: a) 0.120m, b) 1.6s, c) 0.625Hz

Explanation:

a)

From the question, we can see that the maximum displacement from the mean position is 0.120m, this value is the amplitude of the motion.

Hence A = 0.120m

b)

From the point of release to the opposite end, the object has completed half an oscillation ( assuming it went back to it point of release, then we have a complete oscillation), hence the period of the motion is half which is also equals to the time taken to move from the point of release to the opposite end (0.800s)

0.800 = T/2

Where T is the period of motion.

By cross multiplying, we have that

T = 0.800×2

T = 1.6s

c)

There is an inverse relationship between frequency and period.

Where f = 1/T

f = 1/0.16 = 0.625 Hz

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horsena [70]

Answer:

The work is required to stretch it from 8 m to 16 m is 192 N-m

Explanation:

Given that,

Natural length = 8 m

Force F = 12 N

After stretched,

length = 10 m

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x = 10-8=2\ m

Using hook's law

The restoring force is directly proportional to the displacement.

F\propto (-x)

F = -kx

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Now, The value of k is

k = \dfrac{F}{x}

k = \dfrac{12}{2}

k = 6

When stretch the string from 8 m to 16 m.

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

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x = elongation

W=\dfrac{1}{2}\times6\times8\times8

W=192\ N-m

Hence, The work is required to stretch it from 8 m to 16 m is 192 N-m

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3 years ago
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kvv77 [185]
The correct option is B.
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Need help with these two questions pretty please?
lakkis [162]
Hi,

I've found a link that should assist you or answer your question.
http://click.dji.com/ANbvbbP7bwUWtSACp6U_?pm=link&as=0004

Have a nice day!
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If the reference point changes, the ______ of position changes, but the actual position does not change
Rudik [331]

Answer:

Description.

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Motion can be defined as a change in the location (position) of a physical object or body with respect to a reference point.

This ultimately implies that, motion would occur as a result of a change in location (position) of an object with respect to a reference point or frame of reference i.e where it was standing before the effect of an external force.

Mathematically, the motion of an object is described in terms of acceleration, time, distance, speed, velocity, displacement, position, etc.

A reference point refers to a location or physical object from which the motion (movement) of another physical object or body can be determined.

For example, a point on a map is considered to be a reference point because it represents the starting point of an object or person in motion.

Hence, if the reference point changes, the description of position changes, but the actual position does not change.

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baherus [9]

Explanation:

We have,

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The first bright fringe is formed at a distance off 2.9 mm from the center of the pattern. We need to find the wavelength.

For double slit experiment, the fringe width is given by :

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So, wavelength is 433 nm.

6 0
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