Answer:
1. The period doubles
2. The mechanical energy is unchanged
3. The speed is halved
4. The period is unchanged
5. The energy is quadrupled
6. The maximum is speed doubled
Explanation:
1. From Hooke's law we have [tex]T\propto\sqrt{\frac{m}{k} }[/tex]
where T is period, m is mass and k is the spring constant
So If the mass is increased by a factor of 4 then period doubles while k constant
2 According to law of conservation of energy, the energy remains unchanged so therefore the total mechanical energy remain unchanged.
3. From the [tex]w = \sqrt{\frac{k}{m} }[/tex]
where m is mass and w is speed
so we see that mass and speed is inversely proportional therefore if we increase mass speed decreases.
4. The period is independent of amplitude
5. [tex]E=A^{2} [/tex]
where E is energy and A is amplitude
So if the amplitude is doubled the energy is quadrupled.
6. we have the relation
[tex]v_{max}=A [/tex]
where [tex]v_{max}[/tex] is maximum speed and A is amplitude
From the formular we can see they are directly proportional, so if we double amplitude then [tex]v_{max}[/tex] doubles also.
The period would double when the mass of the spring is increased by a factor of four (4).
Since all springs obey Hooke’s law, the period is given by this formula:
[tex]T\alpha \sqrt{\frac{m}{k}}\\\\T = 2\pi \sqrt{\frac{m}{k} }[/tex]
Where:
We can deduce that, the period is directly proportional to mass of the spring. Thus, the period would double when the mass of the spring is increased by a factor of four (4).
In accordance with the law of conservation of energy, the total mechanical energy of the spring would remain the same because energy can neither be created nor destroyed.
The speed of a spring in simple harmonic motion is an angular speed and it is given by this formula:
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
We can deduce that, the angular speed is inversely proportional to mass of the spring. Thus, the angular speed would decrease when the mass of the spring is increased by a factor of four (4).
In simple harmonic motion, the period of a spring is independent of its amplitude.
Mathematically, the the total mechanical energy of a spring is given by this formula:
[tex]E=A^2[/tex]
We can deduce that, the total mechanical energy is directly proportional to the square of amplitude of the spring. Thus, the total mechanical energy would quadruple when the amplitude of the spring is doubled.
Also, the maximum speed is directly proportional to the amplitude of the spring. Thus, the maximum speed would also double when the amplitude of the spring is doubled.
Read more on spring constant here: https://brainly.com/question/25313999
