Transmissibility and Magnification Factor MCQ Quiz - Objective Question with Answer for Transmissibility and Magnification Factor - Download Free PDF

Explanation:

First, let's convert the operational speed from rpm to radians per second:

Operational speed (ω) in radians per second = 2π × (rpm / 60)

Given that the rpm is 450, we have:

ω = 2 × π × (450 / 60)

ω = 2 × π × 7.5

ω = 15π rad/s

Given that π² = 10, we can find π by taking the square root of 10:

π = √10

Now, ω = 15 × √10 rad/s

Next, let's calculate the natural frequency (ωₙ) of the system using the formula:

ωₙ = √(k/m)

Where m is the mass of the refrigerator unit (44 kg) and k is the stiffness of each spring. Since there are 3 springs of the same stiffness, the effective stiffness (kₑ) is:

kₑ = 3k

Therefore, the formula for the natural frequency becomes:

ωₙ = √(3k / 44)

To determine the allowable stiffness, we use the given condition that only 10% of the shaking force is allowed to be transmitted to the supporting structure. This corresponds to a transmissibility ratio (T) of 0.1.

The transmissibility ratio (T) is given by:

\( T_r = \frac{1}{\sqrt{(1 - r^2)^2 + (2 \zeta r)^2}} \)

Where ζ is the damping ratio and r is the frequency ratio (ω/ωₙ).

For a system with negligible damping (ζ ≈ 0), the transmissibility ratio simplifies to:

T = 1 / √[1 + r²]

Given T = 0.1, we have:

0.1 = 1 / √[1 + r²]

Squaring both sides:

(0.1)² = 1 / (1 + r²)

0.01 = 1 / (1 + r²)

1 + r² = 100

r² = 99

r = √99

Substituting r = ω/ωₙ, we get:

√99 = (15 × √10) / ωₙ

ωₙ = (15 × √10) / √99

ωₙ = 15 × √(10/99)

ωₙ = 15 × √(10/99)

ωₙ = 15 × √(10/99)

ωₙ = 15 × (√10 / √99)

ωₙ = 15 × (√10 / √99)

ωₙ = 15 × (1 / √9.9)

ωₙ = 15 × (1 / 3.146)

ωₙ ≈ 4.77 rad/s

Using the formula for natural frequency:

ωₙ = √(3k / 44)

4.77 = √(3k / 44)

Squaring both sides:

(4.77)² = 3k / 44

22.7529 = 3k / 44

3k = 22.7529 × 44

3k = 1001.1276

k = 1001.1276 / 3

k ≈ 333.71 N/m

Converting to N/mm:

k ≈ 333.71 / 1000 N/mm

k ≈ 0.334 N/mm

Concept:

In the vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio.

\(T = \frac{{{F_T}}}{{{F_O}}} = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {\left[ {\left\{1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} \right\}^2+ {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]} }}\)

• When ω/ω_n = 0 ⇒ TR = 1, (independent of ζ)
• When ω/ω_n = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
• When frequency ratio ω/ω_n = √2, then all the curves pass through the point TR = 1 for all values of damping factor ξ.
• When frequency ratio ω/ω_n < √2, then TR > 1 for all values of damping factor ξ. This means that the force transmitted to the foundation through elastic support is greater than the force applied.
• When frequency ratio ω/ω_n > √2, then TR < 1 for all values of damping factor ξ. This shows that the force transmitted through elastic support is less than the applied force. Thus vibration isolation is possible only in the range of ω/ω_n > √2. Here the force transmitted to the foundation increases as the damping is increased.
• If ω/ωn > √2 then transmissibility increases with an increase in damping if ω/ωn < √2 then transmissibility decreases with an increase in damping

Explanation:

The steady-state amplitude for the system:

\(A = \dfrac{{\frac{F}{k}}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2ξ \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

Static deflection (Xst):

\(X_{st}=\dfrac{F}{k}\)

Magnification Factor:

\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2ξ \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)

where \(r=\dfrac{\omega}{\omega_n}\).

Thus M.F = f(r, ξ)

Option 1:

\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)

At r = 1 (resonance)

\(M.F= \dfrac{{1}}{{{2ξ}}}\)

Thus M.F is only a function of the damping ratio i.e. ξ.

Option 2:

\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)

At r = 1 and ξ = 0

\(M.F= \dfrac{{1}}{{{2ξ}}}=\infty\)

The magnification factor is infinite.

Option 3:

\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)

For a given ξ, the value of r at which the M.F is maximum is called ropt.

\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)

When the denominator is minimum, MF will be maximum.

\(\dfrac{d}{dr}\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}}=0\)

\(r_{opt}=\sqrt{1-2\xi^2}\;\)

Concept:

In the vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio.

\(T = \frac{{{F_T}}}{{{F_O}}} = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {\left[ {\left\{1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} \right\}^2+ {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]} }}\)

• When ω/ω_n = 0 ⇒ TR = 1, (independent of ζ)
• When ω/ω_n = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
• When frequency ratio ω/ω_n = √2, then all the curves pass through the point TR = 1 for all values of damping factor ξ.
• When frequency ratio ω/ω_n < √2, then TR > 1 for all values of damping factor ξ. This means that the force transmitted to the foundation through elastic support is greater than the force applied.
• When frequency ratio ω/ω_n > √2, then TR < 1 for all values of damping factor ξ. This shows that the force transmitted through elastic support is less than the applied force. Thus vibration isolation is possible only in the range of ω/ω_n > √2. Here the force transmitted to the foundation increases as the damping is increased.
• If ω/ωn > √2 then transmissibility increases with an increase in damping if ω/ωn < √2 then transmissibility decreases with an increase in damping

Explanation:

The steady-state amplitude for the system:

\(A = \dfrac{{\frac{F}{k}}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2ξ \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

Static deflection (Xst):

\(X_{st}=\dfrac{F}{k}\)

Magnification Factor:

\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2ξ \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)

where \(r=\dfrac{\omega}{\omega_n}\).

Thus M.F = f(r, ξ)

Option 1:

\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)

At r = 1 (resonance)

\(M.F= \dfrac{{1}}{{{2ξ}}}\)

Thus M.F is only a function of the damping ratio i.e. ξ.

Option 2:

\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)

At r = 1 and ξ = 0

\(M.F= \dfrac{{1}}{{{2ξ}}}=\infty\)

The magnification factor is infinite.

Option 3:

\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)

For a given ξ, the value of r at which the M.F is maximum is called ropt.

\(M.F=\dfrac{A}{X_{st}}= \dfrac{{1}}{{\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}} }}\)

When the denominator is minimum, MF will be maximum.

\(\dfrac{d}{dr}\sqrt {{{\left( {1 - {{r}^2}} \right)}^2} + {{\left( {2ξ r} \right)}^2}}=0\)

\(r_{opt}=\sqrt{1-2\xi^2}\;\)

Explanation:

At resonance (ω = ω_n), phase angle ϕ is 90°.

When ω/ω_n ≫ 1; the phase angle is very close to 180°. Here inertia force increases very rapidly and its magnitude is very large.

Concept:

In vibration system, transmissibility is given by

\(\epsilon =\frac{{{F}_{T}}}{{{F}_{O}}}\)

Where F_T = maximum force transmitted to the ground

\(\epsilon =\frac{\sqrt{1+{{\left( 2\xi \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}}}{\sqrt{{{\left( 1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}} \right)}^{2}}+{{\left( \frac{2\xi \omega }{{{\omega }_{n}}} \right)}^{2}}}}\)

Natural frequency \({{\omega }_{n}}=\sqrt{\frac{K}{m}}\)

K = spring stiffness, m = mass

Calculation:

Given, Harmonic excitation force: F(t) = F₀cos(ωt)

F_T = F_O

\(\epsilon =\frac{{{F}_{T}}}{{{F}_{O}}}\)

∴ Transmissibility, ϵ = 1

Also, \(\epsilon =\frac{\sqrt{1+{{\left( 2\xi \frac{w}{{{w}_{n}}} \right)}^{2}}}}{\sqrt{{{\left( 1-{{\left( \frac{w}{{{w}_{n}}} \right)}^{2}} \right)}^{2}}+{{\left( \frac{2\xi w}{{{w}_{n}}} \right)}^{2}}}}=1\)

Let \(\frac{\omega }{{{\omega }_{n}}}=x\)

So, 1 + (2ξ x)² = (1 – x²)² + (2ξ x)²

(1 – x²)² = 1

1 – x² = ± 1

Taking +ve sign x² = 0

∴ Taking –ve sign

x² = 2

Putting value of x:

(ω/ω_n)² = 2

\(\frac{\omega }{{{\omega }_{n}}}=\sqrt{2}\)

∴ ω = √2ω_n

∵ Natural frequency \({{\omega }_{n}}=\sqrt{\frac{K}{m}}\)

Concept:

In the vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio.

\(T = \frac{{{F_T}}}{{{F_O}}} = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {\left[ {\left\{1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} \right\}^2+ {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]} }}\)

• When ω/ω_n = 0 ⇒ TR = 1, (independent of ζ)
• When ω/ω_n = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
• When frequency ratio ω/ω_n = √2, then all the curves pass through the point TR = 1 for all values of damping factor ξ.
• When frequency ratio ω/ω_n < √2, then TR > 1 for all values of damping factor ξ. This means that the force transmitted to the foundation through elastic support is greater than the force applied.
• When frequency ratio ω/ω_n > √2, then TR < 1 for all values of damping factor ξ. This shows that the force transmitted through elastic support is less than the applied force. Thus vibration isolation is possible only in the range of ω/ω_n > √2. Here the force transmitted to the foundation increases as the damping is increased.
• If ω/ωn > √2 then transmissibility increases with an increase in damping if ω/ωn < √2 then transmissibility decreases with an increase in damping

Concept:

Impulse (I) = Change in momentum (ΔP)

Calculation

Given

Mass, m = 5 kg

Stiffness of spring, k = 10 kN/m

At, t = 0 , Mass is at rest u = 0

Force, F = 5 kN is applied for time, t = 10^-4 sec

Using impulse momentum theorem

Impulse (I) = Change in momentum (ΔP)

F × t = m × (v – u)

5 × 10³ × 10^-4  = 1 × (v – 0)

\(v = 0.5~\frac{m}{{sec}}\)

Using energy balance

\(\frac{1}{2}m{v^2} = \frac{1}{2}k{X^2}\)

\(1 \times {0.5^2} = 10 \times {10^3} \times {X^2}\)

X = 5 × 10^-3 = 5 mm

Explanation:

The ratio of maximum displacement of forced vibration to the deflection due to the static force (F) is known as the Magnification factor.

Amplitude of the steady state response

\(A = \frac{{\frac{{{F_0}}}{k}}}{{\sqrt {{{\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }} = \frac{{{X_s}}}{{\sqrt {{{\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

Magnification factor:

\(\frac{A}{{{X_s}}} = \frac{1}{{\sqrt {{{\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

\(MF\; = \;\frac{1}{{\sqrt {{{\left( {1 - \frac{{{\omega ^2}}}{{\omega _n^2}}} \right)}^2}\; + \;{{\left( {\frac{{2{\rm{\zeta \omega }}}}{{{\omega _n}}}} \right)}^2}} }}\)

The ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio ϵ.

\(\varepsilon = \frac{{\sqrt {1 + {{\left( {\frac{{2c\omega }}{{{c_c}.{\omega _n}}}} \right)}^2}} }}{{\sqrt {{{\left( {\frac{{2c\omega }}{{{c_c}.{\omega _n}}}} \right)}^2} + {{\left( {1 - \;\frac{{{\omega ^2}}}{{{\omega _n}^2}}} \right)}^2}\;} }}\)

Logarithmic decrement is defined as the natural logarithm of the amplitude reduction factor. The amplitude reduction factor is the ratio of any two successive amplitudes on the same side of the mean position.

\(\delta = {\log _e}\left( {\frac{{{x_n}}}{{{x_{n + 1}}}}} \right)\)

The amplitude of the vibration is the maximum displacement of a vibrating body from the equilibrium position.

Concept:

In vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio.

\(T = \frac{{{F_T}}}{{{F_O}}} = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]} }}\)

• When ω/ωn = 0 ⇒ TR = 1, (independent of ζ)
• When ω/ωn = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
• When frequency ratio ω/ωn = √2, then all the curves pass through the point TR = 1 for all values of damping factor ξ.
• When frequency ratio ω/ωn < √2, then TR > 1 for all values of damping factor ξ. This means that the force transmitted to the foundation through elastic support is greater than the force applied.
• When frequency ratio ω/ωn > √2, then TR < 1 for all values of damping factor ξ. This shows that the force transmitted through elastic support is less than the applied force. Thus vibration isolation is possible only in the range of ω/ωn > √2. Here the force transmitted to the foundation increases as the damping is increased.

Explanation:

Transmissibility ratio (TR):

In a vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio.

\(TR = \frac{{{F_T}}}{{{F_O}}} = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]} }}\)

• When ω/ω_n = 0 ⇒ TR = 1, (independent of ζ)
• When ω/ωn = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
• When frequency ratio ω/ωn = \(\sqrt{2}\), then all the curves pass through the point TR = 1 for all values of damping factor ξ.
• When frequency ratio ω/ωn < \(\sqrt{2}\), then TR > 1 for all values of damping factor ξ. This means that the force transmitted to the foundation through elastic support is greater than the force applied.
• When frequency ratio ω/ωn > \(\sqrt{2}\), then TR < 1 for all values of damping factor ξ. This shows that the force transmitted through elastic support is less than the applied force. Thus vibration isolation is possible only in the range of ω/ωn > \(\sqrt{2}\). Here the force transmitted to the foundation increases as the damping is increased.

\(\omega = \frac{{2\pi \times 3600}}{{60}} = 377rad/s\)

Natural frequency, \({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{100 \times 1000}}{{250}}} = 20\ rad/s\)

Now, \(r = \frac{\omega }{{{\omega _n}}} = \frac{{377}}{{20}} = 18.85\)

Transmissibility ratio \(\left( {TR} \right) = \frac{{\sqrt {1 + {{\left( {2\xi r} \right)}^2}} }}{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r } \right)}^2}} }}\)

\(= \frac{{\sqrt {1 + {{\left( {2 \times 0.15 \times 18.85} \right)}^2}} }}{{\sqrt {\left[ {1 - {{\left( {18.85} \right)}^2}} \right]^2 + {{\left( {2\times0.15 \times 18.85} \right)}^2}} }}\)

= 0.0162

Concept:

Transmissibility is defined as the ratio of force transmitted to the foundation (FT) to the disturbing force (F).

\(TR = \frac{{{F_T}}}{{{F_o}}} = \frac{{\sqrt {1 + {{\left( {\frac{{2\xi \omega }}{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {{{\left[ {{{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} - 1} \right]}^2} + {{\left( {\frac{{2\xi \omega }}{{{\omega _n}}}} \right)}^2}} }}\)

\({\omega _n} = \sqrt {\frac{k}{m}}\)

For parallel connection of the springs 

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

Calculation:

Given, m = 200 kg

F(t) = 50 cos 5t = Fo cos ωt

Fo = 50, ω = 5, ξ = 0

\({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{{k_1} + {k_2}}}{m}} = \sqrt {\frac{{2k}}{m}} = \sqrt {\frac{{2 \times 10 \times 1000}}{{200}}} = 10\;rad/s\)

\(r = \frac{\omega }{{{\omega _n}}} = \frac{5}{{10}} = 0.5\)

For ξ = 0

\(TR = \frac{{{F_T}}}{{{F_o}}} = \frac{1}{{1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}}} = \frac{1}{{1 - {{\left( {0.5} \right)}^2}}} = 1.333\)

F_T = 1.333 × F_o = 1.333 × 50 = 66.67 N 

Hence the force transmitted to each mount will be,

FT /2 = 66.67/2 = 33.33 N

Explanation:

Adding an additional damper in parallel with the existing damper in an underdamped spring-mass system will increase the damping in the system. This increased damping will lead to a decrease in the amplitude of the system's free oscillations and an increase in the time period of free oscillation.

Concept used:

\(ω_d = \sqrt{1 - ξ^2} . ω_n~\)

where

ω_d = frequency of damped vibration, ωn = nature frequency of vibration, ξ = damping ratio

we know the Time period of oscillation is given by

\(T_d = \frac{2 \pi}{ω_d}\)

Why adding a damper in parallel increases the time period of free oscillation in a single-degree-of-freedom underdamped spring-mass system:
In a spring-mass system, the time period of free oscillation depends on the system's natural frequency and damping ratio. The natural frequency determines the inherent rate at which the system oscillates without any external influence, while the damping ratio affects how quickly the oscillations decay over time.

Adding a damper in parallel to the existing damper effectively increases the total damping coefficient (c) of the system. While the mass (m) and spring constant (k) remain the same, the increase in c directly impacts the damping ratio (ζ) as follows:

⇒ As the damping coefficient increases, the damping ratio also increases.

⇒ ω_d will decrease and as ω_d ↓ T_d ↑ Since they are inversely proportional.

Important Points

Transmissibility  \(ε=\frac{\sqrt{1+\left(\frac{2 ξ \omega}{\omega_{n}}\right)^{2}}}{\sqrt{\left[1-\left(\frac{\omega}{\omega_{n}}\right)^{2}\right]^{2}+\left[\frac{2 ξ \omega}{\omega_{n}}\right]^{2}}}\)

In the underdamped system

ε ↓ if ξ ↑ for \(\frac{\omega}{\omega_n} < \sqrt 2\)

ε ↑ if ξ ↑ for \(\frac{\omega}{\omega_n} > \sqrt 2\)

ε will remain the same if \(\frac{\omega}{\omega_n} = \sqrt 2\)

Hence ε depends on ξ as well as \(\frac{\omega}{\omega_n}\) and there is no information provided related to \(\frac{\omega}{\omega_n}\)

so options (1) and (3) will not be our answer.

Hence the correct answer is option (4) Time period of oscillation will increase.