The drawing shows a top view of a frictionles | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The angular frequencies of the two oscillators are:$\omega_1 = \sqrt{\frac{k}{m_1}} = \sqrt{\frac{1200}{3.0}} = 20 \ rad/s$$\omega_2 = \sqrt{\frac

Vedclass · Accepted Answer

$\frac{3\pi}{40} \ s$

Vedclass · Answer

(C) The angular frequencies of the two oscillators are:$\omega_1 = \sqrt{\frac{k}{m_1}} = \sqrt{\frac{1200}{3.0}} = 20 \ rad/s$$\omega_2 = \sqrt{\frac{k}{m_2}} = \sqrt{\frac{1200}{27}} = \sqrt{\frac{400}{9}} = \frac{20}{3} \ rad/s$Both particles are released from the same initial displacement $A = 10 \ cm$. Their positions as a function of time are $x_1(t) = A \cos(\omega_1 t)$ and $x_2(t) = A \cos(\omega_2 t)$.They are side by side when $x_1(t) = x_2(t)$,which implies $\cos(20t) = \cos(\frac{20}{3}t)$.This occurs when $20t = 2n\pi \pm \frac{20}{3}t$. For the first time after $t=0$,we take the negative sign:$20t = 2\pi - \frac{20}{3}t$$(20 + \frac{20}{3})t = 2\pi$$(\frac{60+20}{3})t = 2\pi$$\frac{80}{3}t = 2\pi$$t = \frac{6\pi}{80} = \frac{3\pi}{40} \ s$.

Similar Questions

The effective spring constant of the two-spring system as shown in the figure will be:

$A$ spring is stretched by $5 \,\,cm$ by a force of $10 \,\,N$. The time period of the oscillations when a mass of $2 \,\,kg$ is suspended by it is: (in $s$)

$A$ body of mass $m$ is attached to a spring which is oscillating with time period $6 \ s$. If the mass of the body is increased by $6 \ kg$,its time period increases by $3 \ s$. Determine the initial mass $m$. (in $kg$)

What will be the force constant of the spring system shown in the figure?

If the period of oscillation of mass $m$ suspended from a spring is $2 \, s$,then the period of mass $4m$ will be .... $s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module