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vagabundo [1.1K]
4 years ago
7

A mass attached to a spring is displaced from its equilibrium position by 5cm and released. The system then oscillates in simple

harmonic motion with a period of 1s. If that same mass–spring system is displaced from equilibrium by 10cm instead, what will its period be in this case?
A mass attached to a spring is displaced from its equilibrium position by and released. The system then oscillates in simple harmonic motion with a period of . If that same mass–spring system is displaced from equilibrium by instead, what will its period be in this case?

A) 0.5s
B) 2s
C) 1s
D) 1.4s
Physics
1 answer:
jok3333 [9.3K]4 years ago
4 0

Answer:

C) 1 s

Explanation:

The period of a mass-spring system is given by the formula:

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass hanging on the spring

k is the spring constant

As we can see from the equation above, the period of the system does NOT depend on the initial amplitude of the oscillation. Therefore, even if the initial amplitude is changed from 5 cm to 10 cm, the period of the system will remain the same, 1 s.

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3 years ago
An object with total mass mtotal = 14.6 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.9 k
zheka24 [161]

Answer: 1) 0. 2) 4.2 Kg. 3) 15.4 m/s 4) 12.9 m/s 5) 0. 6) 3.62 KJ.

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1) Assuming that no external forces act during the collision, total momentum must be conserved. As initially the total mass was at rest, so initial momentum is zero, final momentum of all the system must be 0 also.

2) After the explosion, as mass must be conserved also, the sum of the masses of the three pieces must be equal to the original total mass, so we can write the following:

m₁ + m₂ + m₃ = M = 14.6 Kg = 4.9 Kg + 5.5 Kg + m₃

Solving for m₃, we have:

m₃ = 14.6 Kg - 4.9 Kg -5.5 Kg = 4.2 Kg.

3) and 4)

As momentum is a vector, if it is magnitude must be 0, this means that all his components must be 0 too.

So, we can write two equations, one for the x-component, and other for the y-component, as follows:

pₓ = m₁. v₁ₓ + m₂.v₂ₓ + m₃.v₃ₓ = 0

py = m₁.v₁y + m₂. v₂y + m₃. v₃y =0

Replacing by the values, and solving for v₃ₓ and v₃y, we get:

v₃ₓ = 15.4 m/s

v₃y = 12.9 m/s

v = √(15.4)²+(12.9)² = 20.1 m/s

5) As the center of mass must move as if all the mass were concentrated in this point, and we know that the total momentum must be 0, this tells us that the magnitude of the velocity of the center of mass must be 0 too.

6) As initial kinetic energy is 0, as  the mass was at rest, the increase in the kinetic energy is obtained simply adding the kinetic energy of every piece of mass gained after explosion, as follows:

K = K₁ + K₂ + K₃ = 1/2 (m₁ . v₁² + m₂.v₂² + m₃.v₃²)

Replacing by the values, we get:

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4 0
3 years ago
A 129-kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 65.
LUCKY_DIMON [66]

Answer:

Moment of inertia of the system is 289.088 kg.m^2

Explanation:

Given:

Mass of the platform which is a uniform disk = 129 kg

Radius of the disk rotating about vertical axis = 1.61 m

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Distance from the center of platform = 1.07 m

Mass of the dog on the platform = 27.3 kg

Distance from center of platform = 1.31 m

We have to calculate the moment of inertia.

Formula:

MOI of disk = \frac{MR^2}{2}

Moment of inertia of the person and the dog will be mr^2.

Where m and r are different for both the bodies.

So,

Moment of inertia (I_y_y )  of the system with respect to the axis yy.

⇒ I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g

⇒ I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2

⇒ I_y_y=\frac{129(1.61)^2}{2} +65.7(1.07)^2+27.2(1.31)^2

⇒ I_y_y=289.088\ kg.m^2

The moment of inertia of the system is 289.088 kg.m^2

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