Question

physics plus 1 questions​

Answer 1

Answer:

Given:

Acceleration of SHM object = 0.02 m/s²

At the same instant, the Displacement is 0.08 metres from the mean position.

To find:

Time period of oscillation

Calculation:

Let angular frequency be "ω" and displacement be "x"

So, acc. = 0.02

=> ω²x = 0.02

=> ω²(0.08) = 0.02

=> ω² = 1/4

=> ω = √(¼) = ½

Hence time period be T

∴ T = 2π/ω = 2π/(½) = 4π seconds.

So final answer is 4π seconds.

Answer 2

$\huge{\orange{\underline{\red{\mathtt{GIVEN:-}}}}}$

Acceleration of object =0.02m/s²

displacement of object =8m.

$\huge{\orange{\underline{\red{\mathtt{FIND:-}}}}}$

TIME PERIOD OF OSCILLATION

$\huge{\orange{\underline{\red{\mathtt{NOTE:-}}}}}$

DISPLACEMENT IS DENOTE =x.

angular frequency =w.

$\huge{\orange{\underline{\red{\mathscr{ANSWER:-}}}}}$

W²X=0.02(ACC. 0.02m/s² given)

w²(0.08)=0.02

$\rightarrow$ w²=0.08/0.02

$\rightarrow$ w²=¼

$\rightarrow$ w=√¼

$\rightarrow$ .°.w=½

HENCE,

T PERIOD FORMULA IS 2π/w

$\rightarrow$ T=2π/(½)

$\rightarrow$ T=(2×2)π

$\rightarrow$ .°.T=4π seconds.

$\huge{\orange{\underline{\red{\mathtt{THANKS.}}}}}$