First slide

Applications of SHM

Question

A is a uniform solid cylinder, B is a uniform thin hollow cylinder. Both have same mass and same radius. C is a uniform plank. A, B and C each have same mass equal to m. Friction is sufficient so that there is no slipping anywhere. The time period of small oscillations of the above system is T= $2 π \sqrt{\frac{15 m}{n K}}$ , when the plank is disturbed slightly. Here n is an integer. Find n.

Difficult

A

8
B

7
C

2
D

5

Solution

Angular velocity, $ω$ , of each cylinder will be same at a given instant
Let x be the displacement of the spring from its natural, unstretched position, at a given instant
Energy at this instant:
$\begin{array}{l} E = \frac{1}{2} (\frac{{mR}^{2}}{2} + {mR}^{2}) ω^{2} + \frac{1}{2} ({mR}^{2} + {mR}^{2}) ω^{2} + \frac{1}{2} m (2 R ω)^{2} + \frac{1}{2} {kx}^{2} \\ E = \frac{1}{2} (\frac{{mR}^{2}}{2} + {mR}^{2}) ω^{2} + \frac{1}{2} ({mR}^{2} + {mR}^{2}) ω^{2} + \frac{1}{2} m (2 R ω)^{2} + \frac{1}{2} {kx}^{2} \\ \Rightarrow E = \frac{3 {mv}^{2}}{4} + {mv}^{2} + 2 {mv}^{2} + \frac{1}{2} {kx}^{2} [∵ v = R ω, v = velocity of COM of cylinders] \\ \Rightarrow E = \frac{15 {mv}^{2}}{4} + \frac{1}{2} {kx}^{2} \end{array}$
Energy of system remains constant in the absence of dissipative forces (static friction on cylinders does not do work). $\Rightarrow \frac{d E}{d t} = 0$
$\begin{array}{l} \frac{dE}{dt} = \frac{15}{4} mva + 2 kvx = 0 \\ \Rightarrow a = - \frac{8 k}{15 m} x \\ a α x \Rightarrow SHM \\ T = 2 π \sqrt{\frac{x}{a} ∣} = 2 π \sqrt{\frac{15 m}{8 k}} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App