A particle oscillates in straight line simple harmonically with period 8 second and amplitude 4 2 m . Particle starts from mean position. The ratio of the distance travelled by it in 1 st second of

Question

A particle oscillates in straight line simple harmonically with period 8 second and amplitude 4 2 ​ m . Particle starts from mean position. The ratio of the distance travelled by it in 1 st second of its motion to that in 2 nd second is ( sin 4 5 ∘ = 1/ 2 ​ , sin 2 π ​ = 1 )

TestHub · Accepted Answer

The displacement of a particle executing simple harmonic motion starting from the mean position is given by:

$x (t) = A sin (ω t)$

Given the period $T = 8$ s, the angular frequency is:

$ω = \frac{2 π}{T} = \frac{2 π}{8} = \frac{π}{4}$ rad/s

The amplitude is $A = 42$ m. The displacement at $t = 1$ s is:

$x (1) = 42 sin (\frac{π}{4} \times 1) = 42 \times \frac{1}{2} = 4$ m

The distance travelled in the $1^{st}$ second is:

$d_{1} = x (1) - x (0) = 4 - 0 = 4$ m

The displacement at $t = 2$ s is:

$x (2) = 42 sin (\frac{π}{4} \times 2) = 42 sin (\frac{π}{2}) = 42 \times 1 = 42$ m

The distance travelled in the $2^{nd}$ second is:

$d_{2} = x (2) - x (1) = 42 - 4 = 4 (2 - 1)$ m

The ratio of the distance travelled in the $1^{st}$ second to that in the $2^{nd}$ second is:

$\frac{d _{1}}{d _{2}} = \frac{4}{4 ( 2 - 1 )} = \frac{1}{2 - 1}$

Answer: $1 : (2 - 1)$

Physics - SHM Question with Solution | TestHub

Options:

Doubts & Discussion