Respuesta :
Explanation:
Let [tex]\omega[/tex] is the frequency of oscillation of a block that undergoes simple harmonic motion. The frequency of oscillation is given by :
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]...............(1)
Where
k is the spring constant of the spring
m is the mass of the block
It is clear from equation (1) that the frequency of the oscillation depends on the spring constant and the mass of the block. So, on increasing the spring constant, the block oscillate with a greater frequency.
Answer:
(B) Increase the spring constant
Explanation:
The frequency os oscillation of a block is given by the following formula:
[tex]w = \sqrt{\frac{k}{m}}[/tex]
In which k is the spring constant and m is the mass of the block.
So the answer is either b or c, as the frequency of oscillation is a function of the sping constant and of the mass of the block.
Lets try b first, increasing the spring constant.
For example, with m = 4.
If k = 16, w = 2.
If k = 36, w = 3
So as the sping constant increases, so does the frequency of oscillation.
So the correct answer is:
(B) Increase the spring constant
Why c is wrong?
Suppose we have k = 36.
if m = 4, w = 3
if m = 9, w = 2
So, as the mass increases, the frequency decreases. So c is wrong
