Undamped Free Vibration MCQ Quiz in मराठी - Objective Question with Answer for Undamped Free Vibration - मोफत PDF डाउनलोड करा

Concept:

The natural frequency of a spring-mass system is given by,

\(f_n = \frac {\omega_n}{2\pi}\;\) and \(\omega_n = \sqrt {\frac km}\;\)

where k = spring stiffness (N/m) and m = mass of the body (kg)

Calculation:

Given:

Mass of body (m) = 200 kg, spring stiffness (k) = 80 N/mm = 80 × 10³ N/m

The natural frequency of vibration is:

\(\omega_n = \sqrt {\frac km}\;\)

\(\omega_n = \sqrt {\frac{80\;\times\;10^3}{200}}=\sqrt {400}=20\;rad/s\)

Confusion Points

Based on the units of answer we have to use the formula. If we use \(f_n = \frac {\omega_n}{2\pi}\;\) the answer will be in Hertz or s^-1. 

Mistake Points

Use SI unit while calculating natural frequency.

As per the given units, the answer will not come so first we have to convert it into SI unit.

When we will rotate disc by θ, rod will rotate by angle ϕ as disc is pure rolling on the string

\(\therefore L\phi = r\theta \)

\(\phi = \frac{{r\theta }}{L}\) (∵ \(r = \frac{L}{4}\) as given in question)

\(\therefore \phi = \frac{\theta }{4}\)

Moment of Inertia of disc about point D:

\({I_D} = m{r^2} + \frac{{m{r^2}}}{2} = \frac{3}{2}\;m{r^2}\) (Using parallel axis theorem)

Now, by energy method:

\(\frac{{d{E_{total}}}}{{dt}} = 0\)

E_total = Energy of the disc due to rotation + Energy of the springs

\({E_{total}} = \frac{1}{2}{I_{disc}} \times \;{\dot \theta ^2} + \frac{1}{2}k\;{\left( {2r\theta } \right)^2} + \frac{1}{2}2k{\left( {\frac{L}{2} \cdot \phi } \right)^2}\)

\({E_{total}} = \frac{1}{2}\left( {\frac{3}{2}m{r^2}} \right){\dot \theta ^2} + \frac{1}{2}k\;{\left( {2r\theta } \right)^2} + k{\left( {\frac{L}{2} \cdot \frac{\theta }{4}} \right)^2}\)

\(\frac{{d{E_{total}}}}{{dt}} = 0\)

\(\frac{3}{4}~m{{r}^{2}}\times 2\dot{\theta }~\overset{\ddot{\ }}{\mathop{\theta }}\,+\frac{1}{2}k~\left( 4r \right)\times 2\theta ~\dot{\theta }+\frac{k{{L}^{2}}}{64}\times \left( 2\theta ~\dot{\theta } \right)=0\)

\(\frac{3}{2}m~\left( \frac{{{L}^{2}}}{16} \right)\overset{\ddot{\ }}{\mathop{\theta }}\,+\frac{4{{L}^{2}}}{16}k\theta +\frac{2}{64}k{{L}^{2}}\theta =0\)

\(\left( \frac{3m}{32} \right)\overset{\ddot{\ }}{\mathop{\theta }}\,+\left( \frac{9k}{32} \right)\theta =0\)

\(3m~\overset{\ddot{\ }}{\mathop{\theta }}\,+9k~\theta =0\)

\(\overset{\ddot{\ }}{\mathop{\theta }}\,+\frac{9k}{3m}~\theta =0\Rightarrow ~\overset{\ddot{\ }}{\mathop{\theta }}\,+\frac{3k}{m}~\theta =0\)

Comparing with standard equation

\(\because ~\overset{\ddot{\ }}{\mathop{\theta }}\,+\omega _{n}^{2}~\theta =0\)

\(\omega _{n}^{2}=\frac{3k}{m}\)

\({{\omega }_{n}}=\sqrt{\frac{3k}{m}}\)

Concept:

The deflection at the end of cantilever beam will be same as the deflection produced in spring K_1, if mass m is moved

That implies both the stiffnesses are in parallel.

\(\therefore {\omega _n} = \sqrt {\frac{{{k_1} + {k_2}}}{m}} \)

Explanation:

Equations of simple harmonic motion are as followed

x = A sinωt

ẋ = Aω cosωt

Maximum speed will be when cosωt is maximum i.e 1

⇒ Maximum velocity = Aω

Where A is the Amplitude

ω = natural frequency

f = Frequency = 10 vib/sec

\(\omega = \frac{{2\pi }}{T} = 2\pi f = 20\pi\)

Concept:

When two springs of stiffness k1 and k2 are connected in series, the equivalent stiffness is given by –

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}\)

When two springs of stiffness k1 and k2 are connected in parallel, the equivalent stiffness is given by:

keq = k1 + k2

The natural frequency of the vibration is given by:

\(ω_n=\sqrt{\frac{k_{eq}}{m}}\)

Calculation:

Given:

Let the two springs be k1 and k2.

Natural frequency for first case.

\(ω_{n1}=\sqrt{\frac{k_{1}}{m}}\)

Natural frequency for second case.

\(ω_{n2}=\sqrt{\frac{k_{eq}}{m}}\)

Since the two springs are connected in series.

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}\)

\(\frac{1}{{{k_{eq}}}} = \frac{k_1+k_2}{{{k_1}{k_2}}} \)

\({{{k_{eq}}}} = \frac{{k_1}{k_2}}{k_1+k_2}\)

\(ω_{n2}=\sqrt{\frac{\frac{{k_1}{k_2}}{k_1+k_2}}{m}}\;\)

According to problem

ωn2 = 0.5ωn1

\(\frac{\omega_{n2}}{\omega_{n1}}=\sqrt{\frac{{{k_1}{k_2}}{}}{(k_1+k_2)m}}\times \sqrt{\frac{m}{k_1}}\)

\(\frac{0.5\omega_{n1}}{\omega_{n1}}=\sqrt{\frac{{{}{k_2}}{}}{k_1+k_2}}\;\)

\(2=\sqrt{\frac{{{k_1+k_2}{}}}{k_2}}\;\)

\(4=1+{\frac{{{k_1}{}}}{k_2}}\;\)

\({\frac{{{k_1}{}}}{k_2}}=3\)

Concept:

For a simple pendulum hanging at a point, the time period of the oscillation will be,

\(T = 2\pi \sqrt {\frac{L}{g}} \)

Calculation:

The time period of oscillation of the Bob when it moves from P to Q,

\({T_1} = \frac{1}{2} \times 2\pi \sqrt {\frac{L}{g}} = \frac{T}{2}\)

The time period of oscillation of the Bob when it moves from Q to the next extreme position R as shown in the diagram,

\({T_2} = \frac{1}{2} \times 2\pi \sqrt {\frac{L}{{9g}}} = \frac{T}{6}\)

The complete Time period of oscillation of the Bob when it moves from the first extreme position, P to the next extreme position, R as shown in the diagram,

⇒ T₀ = T₁ + T₂

\( \Rightarrow {{{T}}_0} = \frac{{{T}}}{2} + \frac{{{T}}}{6} \)

\(\Rightarrow T_0= \frac{{2{{T}}}}{3}\)

Concept:

In Simple harmonic motion, we have

Time period, \(T = 2π \sqrt {\frac{m}{k}}\)

where, T = Time period of oscillation, m = mass of the block, k = Spring constant

Calculation:

Given:

We have, m = 4 kg, k = 64 N/m

Since, \(T = 2π \sqrt {\frac{m}{k}}\) placing the given values we get,

\(T = 2π \sqrt {\frac{4}{{64}}} \)

T = π/2

Hence, the period of the resulting motion in seconds is T = π/2.

\({\left( {{f_n}} \right)_1} = \frac{\pi }{2}\sqrt {\frac{{EI}}{{m{L^4}}}} = \frac{{0.5615}}{{\sqrt \delta }},\) fundamental frequency

\({\left( {{f_n}} \right)_2} = {2^2}{\left( {{f_n}} \right)_1}\)

\({\left( {{f_n}} \right)_3} = {3^2}{\left( {{f_n}} \right)_1}\)

\({\left( {{f_n}} \right)_n} = {n^2}{\left( {{f_n}} \right)_1}\)

Two nodes are observed between the support means shaft is vibrating in the third mode.

∴ 2250 = 3²f_n = 9f_n (considering 2 nodes between the ends)

\(\Rightarrow {f_n} = \frac{{2250}}{9} = 250\;rpm\)

Concept:

Whirling frequency of shaft is,

f = n² f_c  

where, n = no. of loop (node + 1), f_c = critical frequency

Since it is simply supported critical speed and first frequency = f_n

Calculation:

Given:

N = 1800 rpm (for 2 node), n = 3, Nc = ?

1800 = 3² × (N_c)

∴ N_c = 200 rpm

Concept:

Let the Transverse natural frequency of rotor is f_n,

\(\frac{1}{{{{\left( {f_n} \right)}^2}}} = \frac{1}{{{{\left( {{f_1}} \right)}^2}}} + \frac{1}{{{{\left( {{f_2}} \right)}^2}}} + \ldots \)

And we know that, 

ω_n = 2πf_n

⇒ \(\frac{2\pi N}{60}=2\pi f_n\)

∴ N = 60f_n

Calculation:

Given:

f₁ = 100 Hz, f₂ =200 Hz  

\(∴ \frac{1}{{{{\left( {{f_n}} \right)}^2}}} = \frac{1}{{{{\left( {100} \right)}^2}}} + \frac{1}{{{{\left( {200} \right)}^2}}}\)

\({\left( {{f_n}} \right)^2} = 8000\)

f_n = 89.44 Hz

Angular velocity

∴ N = 60 f_n

N = 60 × 89.44