Question

The amplitude and time period of a particle
of mass 0.1 kg executing simple harmonic
motion are 1m and 6.28 s, respectively.
Then its angular frequency and
acceleration at a displacement of 0.5 m are
respectively​

Answer 1

${\huge{\bold{\underline{\mathfrak{\orange{QueSTion}}}}}}$

The amplitude and time period of a particle of mass 0.1kg executing simple harmonic motion are 1m and 6.28seconds, respectively. Then it's angular frequency at a displacement of 0.5 respectively?

${\huge{\bold{\underline{\mathfrak{\orange{AnSwEr}}}}}}$

✯$\large{\sf{\underline{\underline{SoMe \: ImporTanT :-}}}}$

$\large\green{\boxed{\sf \omega = \frac{2π}{T} }}$

Here,

•$\large\sf\omega$ =Angular frequency

•$\large\sf\ π$ = 3.14 [approx]

•$\large\sf\ T$ = time period

$\large\green{\boxed{\sf a = -{\omega}^{2}x }}$

Here,

•$\large\sf\omega$ = Angular frequency

•$\large\sf\ a$= Acceleration

•$\large\sf\ x$= displacement

✯$\large{\sf{\underline{\underline{GiVen:-}}}}$

•Amplitude = 1m

•Displacement = 0.5m

•Time period = 6.28s

✯$\large{\sf{\underline{\underline{To \: Find }}}}$

•Angular frequency ($\large\sf\omega$)

•Acceleration ($\large\sf\ a$)

✯$\large{\sf{\underline{\underline{SoluTion:-}}}}$

Now, constitute the given values in formula of Angular frequency

$\large\sf\omega = \frac{2π}{T}$

$\leadsto$$\large\sf\omega = \frac{2 \times\ 3.14}{6.28}$

$\leadsto$$\large\sf\omega = \frac{6.28}{6.28}$

$\red\leadsto$$\large\bf\red{\omega = 1rad}$

Hence,

$\large\sf\ a = -{\omega^2}x$

$\leadsto$$\large\sf\ a =-(1^2) \times\ 0.5$

$\red\leadsto$ $\large\bf\red{a = -0.5m/s^2}$