Question

the period of oscillation of a mass m suspended by an ideal spring is 2 s .if an additional mass of 2 kg be suspended ,the time period is increased by 1 s .find the value of m​

Answer 1

Answer:

Given:

Mass of object = m

Additional mass added = 2 kg

Initial time period = 2 sec.

Final time period = 2 + 1 = 3 sec

To find:

Value of m

Calculation:

Let spring constant be k

$t1 = 2\pi \sqrt{ \frac{m}{k} }$

$= > 2 = 2\pi \sqrt{ \frac{m}{k} }$

Now, new mass = (m + 2)kg

Therefore:

$t2 = 2\pi \sqrt{ \frac{(m + 2)}{k} }$

$3= 2\pi \sqrt{ \frac{(m + 2)}{k} }$

Taking ratio of the 2 Equations :

$\frac{3}{2} = \frac{ \sqrt{(m + 2)} }{ \sqrt{m} }$

Squaring on Both sides:

$= > \frac{9}{4} = \frac{(m + 2)}{m}$

$= > 9m = 4m + 8$

$= > 5m = 8$

$= > m = \frac{8}{5}$

$= > m = 1.6 \: kg$

So mass of object is 1.6 kg.

Answer 2

Hey!

Good question.

✔✔Refer to the attachment.