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Dafna1 [17]
4 years ago
11

You are given a long length of string and an oscillator that can shake one end of the string at any desired frequency. The oscil

lator has a display that indicates the frequency. You are asked to design an experiment to study how the velocity of waves on the string depends on the string's tension. You do not have any way to measure time with sufficient accuracy to help in your investigation.
A. Describe your experimental setup and procedure, including any additional pieces of equipment you would need and the kind of data you would record. Include enough detail that another student could follow and complete the experiment successfully.



B. Describe how you would analyze your data to obtain information about the wave velocity's dependence on tension.

Physics
1 answer:
Ivenika [448]4 years ago
6 0

Answer:

Explanation:

a.

AIM :

TO STUDY HOW VELOCITY OF WAVES ON THE STRING DEPENDS ON THE STRING'S TENSION.

APPARATUS:

Oscillator, long strings , some masses( to create tension in string) and the support ( rectangular wooden piece).

EXPERIMENTAL SETUP:

1. Measure the length of the string and mass of the weights used.

2. Connect one end of string to the oscillator.

3. Place the support below string on table such that the string is in same line without touching table.

4. After the support, the string should hang freely.

5. The other end of string is connected with some small measured masses which should be hanging.

PROCEDURE:

1. Note down the length of string and mass of weights.

2. Adjust the frequency in the oscillator which creates standing waves in the string.

3. Start from lower frequency and note down the lowest frequency at which mild sound is heard or when string forms one loop while oscillating.

4. Calculate the wavelength using of waves using length of string.

5. Calculate the velocity using frequency and wavelength.

6. Calculate linear mass density.

8. Repeat the procedure with different masses.

7. plot a graph with tension in y axis and linear mass density in x axis.

8. Find slope and compare with velocity.

Linear mass density

µ = m/l(kg-1)

tension

T = m x 9.8N

wave length

ƛ = 2L

b.

We can analyze the data by comparing slope of the graph, tension Vs linear mass density with velocity which is constant for constant length.

Write the slope value in terms of value of velocity and find the relationship between velocity and string's tension.

The expected result is

slope = v²

T ∝ V²

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Light is an electromagnetic wave and travels at a speed of 3.00 × 108 m/s. The human eye is most sensitive to yellow-green light
Nimfa-mama [501]

Answer:

f = 5.95 \times 10^{14} Hz

Explanation:

As we know that the frequency of the wave is given as

f = \frac{c}{\lambda}

here we know that

c = 3 \times 10^8 m/s

also we know that

\lambda = 5.04 \times 10^{-7} m

now we have

f = \frac{3 \times 10^8}{5.04 \times 10^{-7}}

f = 5.95 \times 10^{14} Hz

6 0
3 years ago
A snowboarder goes down the hill with a slope of 28° if friction acts on him as he slides down which of the following is the cor
xxMikexx [17]

Answer:

A

Explanation:

All of the frictions are the same, but weight always goes straight down so it can only be A or B. Since they are going down a slope, then the normal force must be sloped. A is the only one out of A and B with a sloped normal force, so it has to be A

6 0
3 years ago
A student was trying to find the relationship between mass and force. He placed four different masses on a table and pulled them
Gwar [14]

Answer:

B. There is a direct proportion between the mass and force listed in the table.

Explanation:

From the table, the values of force increases with increase in the value of mass.

if 5kg=25 N

Finding the contant of proportionality k;

k=25/5=5

thus M=k(F)...........where M is mass in kg and F is force in newton, then

M=5F

This show that for every value of mass, we get the value of Force if we multiply by a contant k=5

This means there is a direct proportionality relation between mass and force in the table.

5 0
3 years ago
Read 2 more answers
A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit
vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

\omega =\sqrt{\dfrac{K}{m}}

The maximum speed v given as

v=\omega A

A=Amplitude

v=\sqrt{\dfrac{K}{m}}\times A

0.32=\sqrt{\dfrac{13.5}{0.75}}\times A

A=0.075 m

A= 0.75 cm

The speed at distance x

v=\omega \sqrt{A^2-x^2}

v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}

v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}

v= 0.21 m/s

5 0
3 years ago
A stone is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stopwatch measures the stone's trajec
olga2289 [7]

Answer:

The height of the cliff is 90.60 meters.

Explanation:

It is given that,

Initial horizontal speed of the stone, u = 10 m/s

Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)

The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s

Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :

h=u't+\dfrac{1}{2}gt^2

h=\dfrac{1}{2}gt^2

h=\dfrac{1}{2}\times 9.8\times (4.3)^2

h = 90.60 meters

So, the height of the cliff is 90.60 meters. Hence, this is the required solution.

5 0
3 years ago
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