3 + Interview Questions in OSCILLATION IN GENERAL AWARENESS Page 1 InterviewSolution

1.	One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of 45º each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.
Answer» Let the liquid in the length dx is at a height x. Its mass = Aρdx PE = (Aρdx)dx The PE of the left column = \(\int^{h_1}_0\) gxdx =A \({[\frac{x^2}{2}]}^{h_1}_0\) = Aρg\(\frac{l^2sin^2\,45^o}{2}\) h₁ = h₂ = l sin 45º where l = length in one arm of the tube. Total PE = Aρgl²sin²45º = \(\frac{Aρgl^2}{2}\) If the change in liquid level in the tube in left side is y, Then length of liquid on left side = l − y And right side = l + y Total PE = Aρg(l − y)² sin²45º + Aρg(l + y)² sin²45º Change in PE = (PE)_F − (PE)_i =\(\frac{Aρg}{2}\) [(l − y)² + (l + y)² − l² ] = \(\frac{Aρg}{2}\)[l² + y² - 2ly + l² + y² + 2ly - l² ] = \(\frac{Aρg}{2}\)[y² + l²] Change in KE = \(\frac{1}{2}\)Aρ2ly² Change in total energy = 0 ∆(PE) + ∆(KE) = 0 \(\frac{Aρg}{2}\) [2y² + l²] + Aρly² = 0 Differentiating both wrt time Aρg[0 + yẏ] + 2aρlẏÿ = 0 2Aρg + 2Aρglÿ = 0 lÿ + gy = 0 Ÿ + \(\frac{g}{l}\)y = 0 Or ÿ = − \(\frac{g}{l}\)y ∴ w² = \(\frac{g}{l}\) or w = \(\sqrt{\frac{g}{l}}\) or T = \(2\pi\sqrt{\frac{l}{g}}\)

1.

One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of 45º each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

Answer»

Let the liquid in the length dx is at a height x.

Its mass = Aρdx

PE = (Aρdx)dx

The PE of the left column = \(\int^{h_1}_0\) gxdx

=A \({[\frac{x^2}{2}]}^{h_1}_0\) = Aρg\(\frac{l^2sin^2\,45^o}{2}\)

h₁ = h₂ = l sin 45º where l = length in one arm of the tube.

Total PE = Aρgl²sin²45º = \(\frac{Aρgl^2}{2}\)

If the change in liquid level in the tube in left side is y,

Then length of liquid on left side = l − y

And right side = l + y

Total PE = Aρg(l − y)² sin²45º + Aρg(l + y)² sin²45º

Change in PE = (PE)_F − (PE)_i =\(\frac{Aρg}{2}\) [(l − y)² + (l + y)² − l² ]

= \(\frac{Aρg}{2}\)[l² + y² - 2ly + l² + y² + 2ly - l² ] = \(\frac{Aρg}{2}\)[y² + l²]

Change in KE = \(\frac{1}{2}\)Aρ2ly²

Change in total energy = 0

∆(PE) + ∆(KE) = 0

\(\frac{Aρg}{2}\) [2y² + l²] + Aρly² = 0

Differentiating both wrt time

Aρg[0 + yẏ] + 2aρlẏÿ = 0

2Aρg + 2Aρglÿ = 0

lÿ + gy = 0

Ÿ + \(\frac{g}{l}\)y = 0

Or ÿ = − \(\frac{g}{l}\)y

∴ w² = \(\frac{g}{l}\) or w = \(\sqrt{\frac{g}{l}}\) or T = \(2\pi\sqrt{\frac{l}{g}}\)

Discussion

2.	Which of the following statements is/are true for a simple harmonic oscillator?(a) Force acting is directly proportional to displacement from the mean position and opposite to it. (b) Motion is periodic.(c) Acceleration of the oscillator is constant.(d) The velocity is periodic.
Answer» (a) Force acting is directly proportional to displacement from the mean position and opposite to it. (b) Motion is periodic. (d) The velocity is periodic.

2.

Which of the following statements is/are true for a simple harmonic oscillator?(a) Force acting is directly proportional to displacement from the mean position and opposite to it. (b) Motion is periodic.(c) Acceleration of the oscillator is constant.(d) The velocity is periodic.

Answer»

(a) Force acting is directly proportional to displacement from the mean position and opposite to it.

(b) Motion is periodic.

(d) The velocity is periodic.

Discussion

3.	A simple harmonic oscillation is represented by the equation:y = 0.40 sin (440 t + 0.61),where y and t are in metre and second respectively. What are the values of (i) amplitude, (ii) angular frequency,. (iii) frequency of oscillations, (iv) time period of oscillations and (v) initial phase?
Answer» The given equation is y = 0.40 sin (440 t + 0.61) Comparing it with the equation of S.H.M y = a sin (ωt + ϕ₀) We have (i) amplitude, a = 0.40 m (ii) Angular frequency, ω = 440 Hz (iii) Frequency of oscillations v = ω/2π = {440}/{2 x 22/7} = 70 Hz (iv) Time period of oscillations, T = 1/v = 1/70 = 0.0143 s. (v) Initial phase angle, ϕ = 0.61 rad

3.

A simple harmonic oscillation is represented by the equation:y = 0.40 sin (440 t + 0.61),where y and t are in metre and second respectively. What are the values of (i) amplitude, (ii) angular frequency,. (iii) frequency of oscillations, (iv) time period of oscillations and (v) initial phase?

Answer»

The given equation is

y = 0.40 sin (440 t + 0.61)

Comparing it with the equation of S.H.M

y = a sin (ωt + ϕ₀)

We have (i) amplitude, a = 0.40 m

(ii) Angular frequency, ω = 440 Hz

(iii) Frequency of oscillations

v = ω/2π = {440}/{2 x 22/7} = 70 Hz

(iv) Time period of oscillations,

T = 1/v = 1/70 = 0.0143 s.

(v) Initial phase angle, ϕ = 0.61 rad

Discussion

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Which of the following statements is/are true for a simple harmonic oscillator?(a) Force acting is directly proportional to displacement from the mean position and opposite to it. (b) Motion is periodic.(c) Acceleration of the oscillator is constant.(d) The velocity is periodic.

A simple harmonic oscillation is represented by the equation:y = 0.40 sin (440 t + 0.61),where y and t are in metre and second respectively. What are the values of (i) amplitude, (ii) angular frequency,. (iii) frequency of oscillations, (iv) time period of oscillations and (v) initial phase?