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blsea [12.9K]
3 years ago
13

A string of density 0.01 kg/m is stretched with a tension of 5N and fixed at both ends. The length of the string is 0.1m. What i

s the first four resonance frequencies in the string?
Physics
1 answer:
Lisa [10]3 years ago
8 0

Answer:

The first four resonance frequency in the string are;

1) 50·√50 Hz

2) 100·√50 Hz

3)150·√50 Hz

4) 200·√50 Hz

Explanation:

The given parameters of the string are;

The density of the string, ρ = 0.01 kg/m

The tension force on the string, T = 5 N

The length of the string, l = 0.1 m

Therefore the mass of the string, m = Length of string × Density of the string

∴ m = 0.01 kg/m × 0.1 m = 0.001 kg

The formula for the fundamental frequency, f₁, is given as follows;

f_1 = \dfrac{\sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{\sqrt{\dfrac{T}{\rho} } }{2 \cdot L}

Where;

f₁ = The fundamental frequency in the string

T = The tension in the string = 5 N

m = The mass of the string = 0.001 kg

L = The length of the string = 0.1 m

ρ = The density of the string = 0.01 kg/m

By plugging in the values of the variables, we have;

f_1 = \dfrac{\sqrt{\dfrac{5}{0.01} } }{2 \times 0.1} = 50 \cdot \sqrt{5}

The first four harmonics are;

f₁, 2·f₁, 3·f₁, 4·f₁

Therefore, we have the first four resonance frequency of the string are as follows;

1 × 50·√50 Hz = 50·√50 Hz

2 × 50·√50 Hz = 100·√50 Hz

3 × 50·√50 Hz = 150·√50 Hz

4 × 50·√50 Hz  = 200·√50 Hz

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3 years ago
a single 1,300 jk cargo car is rolling along a train track at 2.0 m/s when 400 kg of coal is dropped vertically into it. what is
tensa zangetsu [6.8K]

Answer:

1.53 m/s

Explanation:

Given:

Mass of the car (M) = 1300 kg

Mass of the coal (m) = 400 kg

Initial velocity of the car (U) = 2 m/s

Initial velocity of the coal (u) = 0 m/s (Since it is dropped)

When the coal is dropped into the car, then they move with same final velocity.

Let the final velocity be 'v' m/s.

For a closed system, the law of conservation of momentum holds true.

So, initial momentum is equal to final momentum of the car-coal system.

Initial momentum of the car = MU=1300\times 2=2600\ Ns

Initial momentum of the coal = mu=0\ Ns

Total initial momentum is the sum of the above two momentums.

So, total initial momentum = 2600 + 0 = 2600 Ns

Now, final momentum is given as the product of combined mass and final velocity. So,

Final momentum of the system = (M+m)v=(1300+400)v=1700v

Now, from law of conservation of momentum,

Initial momentum = Final momentum

2600=1700v\\\\v=\frac{2600}{1700}\\\\v=1.53\ m/s

Therefore, the final velocity of either of the two masses is same is equal to 1.53 m/s.

4 0
3 years ago
Plzzz help pop asap
Zarrin [17]

Answer:

F = 200 N

Explanation:

Given that,

Mass of the object, m = 20 kg

The acceleration of the object, a = 10 m/s²

We need to find the force exerted by the object. The force acting on the object is given by :

F = ma

So,

F=20\ kg\times 10\ m/s^2\\\\F=200\ N

So, 200 N of force was exerted by the object.

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D. (static equilibrium)

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