Question

in si units the differential equation of an shm is d^2x/dt^2=-36x. find its frequency and period​

Answer 1

Answer:

Given:

Differential equation of SHM is provided as follows :

$\dfrac{ {d}^{2}x }{d {t}^{2} } = - 36x$

To find:

• Frequency
• Time period of SHM

Calculation:

We know that 2nd order derivative of displacement function always gives us Acceleration . So , we can say that :

$\dfrac{ {d}^{2}x }{d {t}^{2} } = - 36x$

$= > acc. = - 36x$

$= > - { \omega}^{2} x = - 36x$

$= > \omega = \sqrt{36}$

$= > \omega = 6 \: hz$

So time period be T

$T = \dfrac{2\pi}{ \omega} = \dfrac{2\pi}{6} = \dfrac{\pi}{3} sec$

Let frequency be f

$f = \dfrac{1}{T} = \dfrac{3}{\pi} \: hz$

Answer 2

Differential equation of SHM is provided as follows :

\dfrac{ {d}^{2}x }{d {t}^{2} } = - 36x

dt

2

d

2

x

=−36x

To find:

Frequency

Time period of SHM

Calculation:

We know that 2nd order derivative of displacement function always gives us Acceleration . So , we can say that :

\dfrac{ {d}^{2}x }{d {t}^{2} } = - 36x

dt

2

d

2

x

=−36x

= > acc. = - 36x=>acc.=−36x

= > - { \omega}^{2} x = - 36x=>−ω

2

x=−36x

= > \omega = \sqrt{36}=>ω=

36

= > \omega = 6 \: hz=>ω=6hz

So time period be T

T = \dfrac{2\pi}{ \omega} = \dfrac{2\pi}{6} = \dfrac{\pi}{3} secT=

ω

2π

=

6

2π

=

3

π

sec

Let frequency be f

f = \dfrac{1}{T} = \dfrac{3}{\pi} \: hzf=

T

1

=

π

3

hz