Question

If the mass is halved what will be the percentage decrease in the time period of a mass spring system?​

Answer 1

Answer:

Given:

Mass is halved in a spring system.

To find:

% decrease in mass-spring system.

Formulas used:

$time \: period = 2\pi \sqrt{ \frac{m}{k} }$

Calculation:

Let initial mass be "m" and initial time period be t1.

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

Now , new mass be m/2 and new time period be t2.

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

$= > t2= 2\pi \sqrt{ \frac{m}{2k} }$

$= > t2= \frac{1}{ \sqrt{2} }( 2\pi \sqrt{ \frac{m}{k} } )$

$= > t2 = \frac{1}{ \sqrt{2} } (t1)$

So, decrease in time period

$= t1 - t2$

$= t1 - \frac{1}{ \sqrt{2} } (t1)$

$( \frac{ \sqrt{2} - 1}{ \sqrt{2} } )t1$

Therefore % decrease

$= \frac{ \Delta \: t}{t1} \times 100\%$

$= \frac{ \sqrt{2} - 1 }{ \sqrt{2} } \times 100\%$

$= \frac{0.414}{1.414} \times 100\%$

$= 0.292 \times 100\%$

$= 29.2\%$

So final answer :

$\boxed {\red{\% \: change \: = 29.2\%}}$

Answer 2

Answer:

${\boxed{\bold{Decrease\:in\:time\:period=29.2\%}}}$

Explanation:

Given,

• Mass is halved in spring system.
• Percentage decrease in time period=?

We know that,

${\boxed{\bold{Time\:period=2\pi</h3><h3>\sqrt{\frac{m}{k}}}}}$

Let new time period be t2 and mass be m/2.

${\underline{\underline{\bold{According\:to\:question:}}}}$

$\rightarrow\tt t_2 =2\pi \sqrt{\frac{\frac{m}{2}}{k}} \\\rightarrow\tt t_2 = 2\pi \sqrt{\frac{m}{2k}} \\\rightarrow\tt t_2 = \frac{1}{\sqrt{2}} (2\pi \sqrt{\frac{m}{k}}) \\\rightarrow\tt t_2 = \frac{1}{\sqrt{2}} \times t_1 \\\tt *Percentage\:decrease\:in\:time\:period: \\\rightarrow\tt t_1 - t_2 = t_1 - \frac{1}{\sqrt{2}} \times t_1 \\\rightarrow\tt t_1 - \frac{t_1}{\sqrt{2}} \\\rightarrow\tt \frac{\sqrt{2}t_1 - t_1}{\sqrt{2} } \\\rightarrow\tt \frac{\sqrt{2} - 1}{\sqrt{2}} \times 100\% \\\rightarrow\tt \frac{0.4142}{1.4142}\times 100\% \\\\\rightarrow{\boxed{\tt{ 29.2\% }}}$