Question

A block attached at the end of a spring undergoes simple harmonic motion with frequency of oscillation ω.

Answer 1

Answer:

(B) Increase the spring constant

Explanation:

The frequency os oscillation of a block is given by the following formula:

$w = \sqrt{\frac{k}{m}}$

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

Answer 2

Explanation:

Let $\omega$ is the frequency of oscillation of a block that undergoes  simple harmonic motion. The frequency of oscillation is given by :

$\omega=\sqrt{\dfrac{k}{m}}$...............(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.