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11.4 Simple Harmonic Motion (SHM), Pendulum, Mass-Spring & Resonance

Key Takeaways

  • Simple Harmonic Motion (SHM) is defined by the acceleration differential relation a=−ω2xa = -\omega^2 x, where acceleration is directly proportional to displacement and continuously directed toward the mean equilibrium position.

  • For a mass-spring system, the period T=2πmkT = 2\pi\sqrt{\frac{m}{k}} is governed by inertial mass and elastic stiffness; cutting a spring into nn equal segments multiplies the spring constant of each segment by nn (k′=nkk' = nk).

  • A simple pendulum has period T=2πlgT = 2\pi\sqrt{\frac{l}{g}} independent of bob mass and angular amplitude (for small angles θ≤10∘\theta \le 10^\circ); a second's pendulum has T=2 sT = 2\text{ s} and a precise length of l≈0.992 ml \approx 0.992\text{ m} on Earth.

  • Total mechanical energy in SHM is invariant (Etotal=12kx02E_{\text{total}} = \frac{1}{2}kx_0^2); kinetic and potential energies equalize (K=U=12EtotalK = U = \frac{1}{2}E_{\text{total}}) at displacement x=x02≈0.707x0x = \frac{x_0}{\sqrt{2}} \approx 0.707 x_0, and oscillate at twice the system frequency (fE=2ff_E = 2f).

  • Resonance occurs when the driving frequency matches the natural frequency (fdrive=f0f_{\text{drive}} = f_0), resulting in maximal energy transfer and peaked amplitudes whose sharpness is inversely governed by system damping.

Last updated: October 2026

Simple Harmonic Motion (SHM), Pendulum, Mass-Spring & Resonance

Oscillatory and vibrational mechanics, detailed in of the Intermediate Physics Part I syllabus, represent one of the most conceptually rich and mathematically tested areas of the UET ECAT. Simple Harmonic Motion (SHM) describes periodic back-and-forth motion governed by a linear restoring force directed toward a stable equilibrium state. In engineering entrance exams, problems frequently test analytical phase differences, instantaneous kinetic and potential energy partitions, spring combinations, effective pendulum lengths across varying gravitational fields, and resonance phenomena. Developing rapid algebraic fluency with SHM formulas without relying on numerical calculators is essential for exam pacing.


1. Kinematics and Fundamental Definition of SHM

The Fundamental Dynamical Definition

A mechanical system undergoes Simple Harmonic Motion (SHM) if and only if its acceleration aa is directly proportional to its displacement xx from the mean equilibrium position and is always directed toward that mean position:

a∝−x  ⟹  a=−ω2xa \propto -x \implies a = -\omega^2 x

Where:

  • xx is the instantaneous displacement from equilibrium (meters, m\text{m}).
  • ω\omega is the circular angular frequency of oscillation (radians per second, rad⋅s−1\text{rad}\cdot\text{s}^{-1}).
  • The negative sign indicates that acceleration is invariably directed opposite to displacement—namely, pointing back toward x=0x = 0.

By Newton's second law (F=maF = m a):

F=m(−ω2x)=−mω2x=−kxF = m(-\omega^2 x) = -m\omega^2 x = -k x

Where k=mω2k = m\omega^2 is the restoring spring stiffness constant (N⋅m−1\text{N}\cdot\text{m}^{-1}), confirming Hooke's law as the dynamical origin of SHM.

Equations of Kinematics in SHM

Representing SHM as the projection of uniform circular motion of reference radius x0x_0 (amplitude) onto a diameter yields the standard kinematic wave functions:

  1. Instantaneous Displacement (xx):

x(t)=x0sin⁡(ωt+ϕ)x(t) = x_0 \sin(\omega t + \phi)

If timing starts at the mean position (x=0x=0 at t=0t=0), initial phase ϕ=0  ⟹  x(t)=x0sin⁡(ωt)\phi = 0 \implies x(t) = x_0 \sin(\omega t). If timing starts at maximum positive amplitude (x=x0x = x_0 at t=0t=0), ϕ=π2  ⟹  x(t)=x0cos⁡(ωt)\phi = \frac{\pi}{2} \implies x(t) = x_0 \cos(\omega t).

  1. Instantaneous Velocity (vv): Differentiating displacement with respect to time:

v(t)=dxdt=ωx0cos⁡(ωt+ϕ)v(t) = \frac{dx}{dt} = \omega x_0 \cos(\omega t + \phi)

Using the identity cos⁡θ=±1−sin⁡2θ=±1−(x/x0)2\cos\theta = \pm\sqrt{1 - \sin^2\theta} = \pm\sqrt{1 - (x/x_0)^2}:

v(x)=±ωx02−x2v(x) = \pm \omega \sqrt{x_0^2 - x^2}

  • At the mean position (x=0x = 0): Speed attains its maximum value: vmax=ωx0v_{\text{max}} = \omega x_0.
  • At the extreme positions (x=±x0x = \pm x_0): Speed drops to zero: v=0v = 0.
  1. Instantaneous Acceleration (aa): Differentiating velocity with respect to time:

a(t)=dvdt=−ω2x0sin⁡(ωt+ϕ)=−ω2xa(t) = \frac{dv}{dt} = -\omega^2 x_0 \sin(\omega t + \phi) = -\omega^2 x

  • At the mean position (x=0x = 0): Acceleration is zero: a=0a = 0.
  • At the extreme positions (x=±x0x = \pm x_0): Acceleration attains maximum magnitude: amax=ω2x0a_{\text{max}} = \omega^2 x_0.

Phase Relationships Between Kinematic Variables

A central focus of ECAT multiple-choice questions is the phase angle difference between SHM kinematic variables:

  • Displacement vs Velocity: Velocity leads displacement by a phase angle of π2 rad\frac{\pi}{2}\text{ rad} (90∘90^\circ).
  • Velocity vs Acceleration: Acceleration leads velocity by a phase angle of π2 rad\frac{\pi}{2}\text{ rad} (90∘90^\circ).
  • Displacement vs Acceleration: Acceleration and displacement are in exact phase opposition, differing by a phase angle of π rad\pi\text{ rad} (180∘180^\circ).

2. Mass-Spring Systems

Horizontal and Vertical Oscillators

Consider an ideal block of mass mm attached to a massless spring of stiffness constant kk.

  • Restoring Force: F=−kxF = -k x.
  • Equating to Newton's law: ma=−kx  ⟹  a=−(km)xm a = -k x \implies a = -\left(\frac{k}{m}\right)x.

Comparing with a=−ω2xa = -\omega^2 x confirms that the angular frequency is:

ω=km\omega = \sqrt{\frac{k}{m}}

From ω=2πf=2πT\omega = 2\pi f = \frac{2\pi}{T}, the time period TT and frequency ff are:

T=2πmk,f=12πkmT = 2\pi \sqrt{\frac{m}{k}}, \qquad f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

Crucial ECAT Observations:

  • The time period depends solely on inertial mass mm and elastic stiffness kk.
  • TT is completely independent of amplitude x0x_0 (isochronous property).
  • For a vertical mass-spring system, gravity stretches the spring to a new equilibrium extension xeq=mgkx_{\text{eq}} = \frac{mg}{k}. The period about this new equilibrium remains completely unchanged: T=2πmk=2πxeqgT = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{x_{\text{eq}}}{g}}.

Combination of Springs

When multiple springs are coupled together, their effective spring constant keqk_{\text{eq}} dictates the resulting time period:

  1. Series Combination (End-to-End):
    • Each spring experiences the identical tension force FF.
    • Total displacement is the sum of individual extensions: xtotal=x1+x2x_{\text{total}} = x_1 + x_2.
    • Since x=F/kx = F/k, Fkeq=Fk1+Fk2\frac{F}{k_{\text{eq}}} = \frac{F}{k_1} + \frac{F}{k_2}:

1keq=1k1+1k2  ⟹  keq=k1k2k1+k2\frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2} \implies k_{\text{eq}} = \frac{k_1 k_2}{k_1 + k_2}

For two identical springs (k1=k2=kk_1 = k_2 = k):

keq=k2  ⟹  Tseries=2πmk/2=2T0k_{\text{eq}} = \frac{k}{2} \implies T_{\text{series}} = 2\pi\sqrt{\frac{m}{k/2}} = \sqrt{2} T_0

  1. Parallel Combination (Side-by-Side):
    • Both springs undergo identical displacement xx.
    • Total restoring force is the sum of individual spring forces: Ftotal=F1+F2=k1x+k2xF_{\text{total}} = F_1 + F_2 = k_1 x + k_2 x:

keq=k1+k2k_{\text{eq}} = k_1 + k_2

For two identical springs (k1=k2=kk_1 = k_2 = k):

keq=2k  ⟹  Tparallel=2πm2k=T02k_{\text{eq}} = 2k \implies T_{\text{parallel}} = 2\pi\sqrt{\frac{m}{2k}} = \frac{T_0}{\sqrt{2}}

  1. Cutting a Spring:
    • The spring constant of a uniform spring is inversely proportional to its resting length: k∝1L  ⟹  kL=constantk \propto \frac{1}{L} \implies k L = \text{constant}.
    • If a spring of stiffness kk and length LL is cut into nn equal segments, each segment has length L′=L/nL' = L/n, so its new spring constant is:

k′=nkk' = n k


3. The Simple Pendulum

Dynamical Derivation for Small Angular Displacements

A simple pendulum consists of a small heavy bob of mass mm suspended by a light, inextensible string of length ll. Displacing the bob by a small angle θ\theta from the vertical:

  • Weight mgmg resolves into two components: mgcos⁡θmg\cos\theta along the string (balanced by string tension TstringT_{\text{string}}) and mgsin⁡θmg\sin\theta perpendicular to the string, acting as the restoring force toward equilibrium:

Frestoring=−mgsin⁡θF_{\text{restoring}} = -m g \sin\theta

For small angular amplitudes (typically θ≤10∘≈0.175 rad\theta \le 10^\circ \approx 0.175\text{ rad}), the small-angle approximation sin⁡θ≈θ\sin\theta \approx \theta applies. Expressing θ=xl\theta = \frac{x}{l} (where xx is linear arc displacement):

F=−mg(xl)=−(mgl)xF = -m g \left(\frac{x}{l}\right) = -\left(\frac{m g}{l}\right) x

By Newton's second law (F=maF = m a):

ma=−(mgl)x  ⟹  a=−(gl)xm a = -\left(\frac{m g}{l}\right) x \implies a = -\left(\frac{g}{l}\right) x

Comparing with the definition a=−ω2xa = -\omega^2 x:

ω=gl\omega = \sqrt{\frac{g}{l}}

The time period TT and frequency ff are therefore:

T=2πlg,f=12πglT = 2\pi \sqrt{\frac{l}{g}}, \qquad f = \frac{1}{2\pi}\sqrt{\frac{g}{l}}

Key Laws of the Simple Pendulum for ECAT

  1. Law of Length: T∝lT \propto \sqrt{l}. If pendulum length is quadrupled (4l4l), the time period doubles (2T2T).
  2. Law of Gravity: T∝1gT \propto \frac{1}{\sqrt{g}}. If gravitational acceleration decreases (e.g., atop a high mountain or on the Moon where gm=g/6g_m = g/6), the period increases (Tmoon=6T≈2.45TT_{\text{moon}} = \sqrt{6} T \approx 2.45 T), causing the clock to run slow.
  3. Law of Mass Independence: TT is strictly independent of the mass mm and material of the bob. Replacing an aluminum bob with a lead bob of identical dimensions leaves TT unaltered.
  4. Law of Isochronism: TT is independent of angular amplitude θ\theta, provided θ\theta remains small (θ≤10∘\theta \le 10^\circ).

The Second's Pendulum

A second's pendulum is defined as a pendulum whose half-period (time for one single tick from one extreme to the other) is exactly 1 second1\text{ second}, corresponding to a full vibration time period of:

T=2.0 sT = 2.0\text{ s}

To find its length on Earth's surface where g=9.8 m⋅s−2g = 9.8\text{ m}\cdot\text{s}^{-2}:

T=2πlg  ⟹  2.0=2πl9.8  ⟹  1=πl9.8T = 2\pi\sqrt{\frac{l}{g}} \implies 2.0 = 2\pi\sqrt{\frac{l}{9.8}} \implies 1 = \pi\sqrt{\frac{l}{9.8}}

l=9.8π2≈9.809.87≈0.992 m=99.2 cm≈1.0 ml = \frac{9.8}{\pi^2} \approx \frac{9.80}{9.87} \approx 0.992\text{ m} = 99.2\text{ cm} \approx 1.0\text{ m}

Pendulum in Non-Inertial Frames (Accelerated Systems)

If the suspension pivot of a simple pendulum undergoes linear acceleration a0\mathbf{a}_0, an effective gravitational acceleration geff=g−a0\mathbf{g}_{\text{eff}} = \mathbf{g} - \mathbf{a}_0 must be used:

  • Elevator accelerating upward with +a+a: geff=g+a  ⟹  T=2πlg+ag_{\text{eff}} = g + a \implies T = 2\pi\sqrt{\frac{l}{g+a}} (Period decreases, clock runs fast).
  • Elevator accelerating downward with −a-a: geff=g−a  ⟹  T=2πlg−ag_{\text{eff}} = g - a \implies T = 2\pi\sqrt{\frac{l}{g-a}} (Period increases, clock runs slow).
  • Freely Falling Elevator (a=ga = g): geff=g−g=0  ⟹  T=2πl0→∞g_{\text{eff}} = g - g = 0 \implies T = 2\pi\sqrt{\frac{l}{0}} \to \infty (Pendulum ceases oscillating).
  • Geostationary / Orbiting Satellite: Apparent weightlessness (geff=0g_{\text{eff}} = 0) means T→∞T \to \infty; a pendulum cannot tick inside an orbiting spacecraft.

Comparison of Classical Oscillating Systems

Parameter / FeatureHorizontal Mass-Spring SystemVertical Mass-Spring SystemSimple PendulumSecond's Pendulum
Restoring AgentSpring elasticity (F=−kxF = -kx)Net elastic force (F=−kxF = -kx)Tangential gravity component (F=−mgsin⁡θF = -mg\sin\theta)Tangential gravity component (F=−mgsin⁡θF = -mg\sin\theta)
Inertial FactorMass mmMass mmLength ll (independent of mm)Length l≈0.992 ml \approx 0.992\text{ m} (at g=9.8 m⋅s−2g = 9.8\text{ m}\cdot\text{s}^{-2})
Restoring / Elastic FactorSpring constant kkSpring constant kkGravitational acceleration ggGravitational acceleration gg
Angular Frequency (ω\omega)km\sqrt{\frac{k}{m}}km\sqrt{\frac{k}{m}}gl\sqrt{\frac{g}{l}}π rad⋅s−1\pi\text{ rad}\cdot\text{s}^{-1}
Time Period (TT)2πmk2\pi\sqrt{\frac{m}{k}}2πmk=2πxeqg2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{x_{\text{eq}}}{g}}2πlg2\pi\sqrt{\frac{l}{g}}Exactly 2.00 s2.00\text{ s}
Frequency (ff)12πkm\frac{1}{2\pi}\sqrt{\frac{k}{m}}12πkm\frac{1}{2\pi}\sqrt{\frac{k}{m}}12πgl\frac{1}{2\pi}\sqrt{\frac{g}{l}}0.50 Hz0.50\text{ Hz}
Dependence on AmplitudeIndependent (isochronous)Independent (isochronous)Independent for small θ≤10∘\theta \le 10^\circIndependent for small θ≤10∘\theta \le 10^\circ
Dependence on Gravity (gg)Independent (gg has no effect)Independent (TT invariant; gg shifts xeqx_{\text{eq}})Inversely proportional (T∝1/gT \propto 1/\sqrt{g})Inversely proportional (T∝1/gT \propto 1/\sqrt{g})
Behavior in Free Fall (geff=0g_{\text{eff}} = 0)Oscillates normally (TT unchanged)Oscillates normally (TT unchanged)Ceases oscillating (T→∞T \to \infty)Ceases oscillating (T→∞T \to \infty)

4. Energy Conservation and Partition in SHM

In an ideal undamped simple harmonic oscillator, total mechanical energy continuously transfers between kinetic energy (KK) and potential energy (UU), while their sum remains strictly constant.

Instantaneous Potential Energy (UU)

The work done against the restoring force F=kxF = kx to displace the mass from 0 to xx is stored as elastic potential energy:

U=∫0xkx′dx′=12kx2U = \int_0^x kx' dx' = \frac{1}{2} k x^2

Substituting x=x0sin⁡(ωt)x = x_0 \sin(\omega t):

U(t)=12kx02sin⁡2(ωt)U(t) = \frac{1}{2} k x_0^2 \sin^2(\omega t)

  • At the mean position (x=0x = 0): Umin=0U_{\text{min}} = 0.
  • At the extreme positions (x=±x0x = \pm x_0): Umax=12kx02U_{\text{max}} = \frac{1}{2} k x_0^2.

Instantaneous Kinetic Energy (KK)

The kinetic energy of mass mm moving with speed v=ωx02−x2v = \omega\sqrt{x_0^2 - x^2} is:

K=12mv2=12mω2(x02−x2)K = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 (x_0^2 - x^2)

Recalling that k=mω2k = m\omega^2:

K=12k(x02−x2)K = \frac{1}{2} k (x_0^2 - x^2)

Substituting v=ωx0cos⁡(ωt)v = \omega x_0 \cos(\omega t):

K(t)=12kx02cos⁡2(ωt)K(t) = \frac{1}{2} k x_0^2 \cos^2(\omega t)

  • At the mean position (x=0x = 0): Kmax=12kx02K_{\text{max}} = \frac{1}{2} k x_0^2.
  • At the extreme positions (x=±x0x = \pm x_0): Kmin=0K_{\text{min}} = 0.

Total Mechanical Energy (EtotalE_{\text{total}})

Summing kinetic and potential energies at any arbitrary displacement xx:

Etotal=K+U=12k(x02−x2)+12kx2=12kx02=12mω2x02=constantE_{\text{total}} = K + U = \frac{1}{2} k (x_0^2 - x^2) + \frac{1}{2} k x^2 = \frac{1}{2} k x_0^2 = \frac{1}{2} m \omega^2 x_0^2 = \text{constant}

Total mechanical energy is proportional to the square of the amplitude (Etotal∝x02E_{\text{total}} \propto x_0^2) and the square of the frequency (Etotal∝f2E_{\text{total}} \propto f^2).

Specific Energy Partition Locations

  1. Equipartition Location (K=UK = U):

12k(x02−x2)=12kx2  ⟹  x02−x2=x2  ⟹  2x2=x02\frac{1}{2} k (x_0^2 - x^2) = \frac{1}{2} k x^2 \implies x_0^2 - x^2 = x^2 \implies 2x^2 = x_0^2

x=±x02≈±0.707x0x = \pm \frac{x_0}{\sqrt{2}} \approx \pm 0.707 x_0

Kinetic and potential energies are equal at approximately 70.7%70.7\% of maximum displacement.

  1. Midway Position (x=x0/2x = x_0 / 2):

    • Potential energy: U=12k(x0/2)2=14(12kx02)=14Etotal=25%U = \frac{1}{2} k (x_0/2)^2 = \frac{1}{4} \left(\frac{1}{2} k x_0^2\right) = \frac{1}{4} E_{\text{total}} = 25\%.
    • Kinetic energy: K=Etotal−U=34Etotal=75%K = E_{\text{total}} - U = \frac{3}{4} E_{\text{total}} = 75\%.
    • Ratio: KU=3\frac{K}{U} = 3.
  2. Frequency of Energy Oscillations:

    • Because sin⁡2(ωt)=1−cos⁡(2ωt)2\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2} and cos⁡2(ωt)=1+cos⁡(2ωt)2\cos^2(\omega t) = \frac{1 + \cos(2\omega t)}{2}, both potential and kinetic energies oscillate with angular frequency ωE=2ω\omega_E = 2\omega and frequency fE=2ff_E = 2f.
    • In one complete cycle of motion, kinetic energy drops to zero twice and peaks twice.

5. Free, Damped, Forced Oscillations and Resonance

Free vs Damped Oscillations

  1. Free Oscillations: An isolated system oscillates exclusively under internal restoring forces without external excitation or dissipative resistance. It vibrates at its natural frequency f0f_0 with permanent constant amplitude.
  2. Damped Oscillations: Real oscillators encounter dissipative viscous drag or mechanical friction, transferring mechanical energy into thermal energy. The equation of motion includes a damping term bdxdtb\frac{dx}{dt}:

md2xdt2+bdxdt+kx=0m\frac{d^2 x}{dt^2} + b\frac{dx}{dt} + kx = 0

Three Damping Regimes:

  • Underdamping (b<2kmb < 2\sqrt{km}): The system oscillates with an amplitude decaying exponentially over time: x(t)=x0e−γtcos⁡(ω′t)x(t) = x_0 e^{-\gamma t} \cos(\omega' t).
  • Critical Damping (b=2kmb = 2\sqrt{km}): The system returns to equilibrium in the minimum possible time without completing a single oscillation. Practical applications: shock absorbers in automobiles, dead-beat pointer galvanometers.
  • Overdamping (b>2kmb > 2\sqrt{km}): The system slowly creeps back to equilibrium without oscillating, hindered by heavy resistance (e.g., deep oil dashpots, automatic door closers).

Forced Oscillations and Resonance

When an oscillating system is driven by a periodic external force F(t)=F0cos⁡(ωdrivet)F(t) = F_0 \cos(\omega_{\text{drive}} t), it initially exhibits transient behavior and subsequently settles into steady-state motion vibrating at the driving frequency ωdrive\omega_{\text{drive}}.

Resonance Phenomenon

Resonance occurs when the driving frequency precisely matches the natural undamped frequency of the oscillating system:

fdrive=f0f_{\text{drive}} = f_0

Under this resonance condition:

  • The rate of energy absorption from the external driver reaches its absolute maximum.
  • The oscillation amplitude reaches a pronounced peak.
  • The phase difference between the driving force and displacement is exactly π2 rad\frac{\pi}{2}\text{ rad} (90∘90^\circ).

Sharpness of Resonance

The profile of the amplitude vs driving frequency response curve depends on the degree of damping:

  • Small Damping (Low Friction): Yields a high, narrow, sharp resonance peak. The system is highly selective (e.g., LC radio tuning circuits picking up a distinct transmission frequency).
  • Large Damping (High Friction): Yields a low, broad, flat resonance peak. The amplitude remains modest over a wide frequency band.

Practical Engineering and Physical Manifestations of Resonance

  1. Radio and Television Tuning: Variable air capacitors or varactor diodes alter capacitance CC, shifting the electrical natural frequency f0=12πLCf_0 = \frac{1}{2\pi\sqrt{LC}} until it resonates with a targeted broadcast station's carrier wave.
  2. Microwave Cooking: Microwave ovens radiate electromagnetic waves at 2.45 GHz2.45\text{ GHz}. Textbooks often cite this as resonance with water molecules; strictly, food is heated by dielectric loss as polar water molecules repeatedly try to realign with the oscillating field.
  3. Magnetic Resonance Imaging (MRI): Strong magnetic fields align hydrogen nuclear spins, which are excited by resonant radiofrequency pulses to produce cross-sectional clinical imaging.
  4. Destructive Mechanical Resonance: Marching soldiers break step across suspension bridges to prevent periodic footfalls from matching the bridge's natural fundamental vibration modes, which could induce destructive high-amplitude structural oscillations.
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Continuous Energy Exchange Cycle in Simple Harmonic Motion
Test Your Knowledge

A particle executes Simple Harmonic Motion with amplitude x0x_0. At what displacement from the mean equilibrium position is its instantaneous kinetic energy exactly equal to its instantaneous potential energy?

A

±x02\pm \frac{x_0}{\sqrt{2}}

B

±x02\pm \frac{x_0}{2}

C

±x04\pm \frac{x_0}{4}

D

±3x02\pm \frac{\sqrt{3}x_0}{2}

Test Your Knowledge

A uniform spring of spring constant kk is cut into three equal segments. The three identical pieces are then connected side-by-side in parallel to support a mass mm. What is the time period of vertical oscillation of this combined system?

A

2πmk2\pi\sqrt{\frac{m}{k}}

B

2π3mk2\pi\sqrt{\frac{3m}{k}}

C

2π3m3k\frac{2\pi}{3}\sqrt{\frac{m}{3k}}

D

2π3mk\frac{2\pi}{3}\sqrt{\frac{m}{k}}

Test Your Knowledge

A simple pendulum has a time period of 2.0 s2.0\text{ s} on Earth's surface where g=9.8 m⋅s−2g = 9.8\text{ m}\cdot\text{s}^{-2}. If the pendulum is placed inside an elevator accelerating vertically upward at a=2.45 m⋅s−2=g4a = 2.45\text{ m}\cdot\text{s}^{-2} = \frac{g}{4}, what is its new time period of oscillation?

A

2.50 s2.50\text{ s}

B

1.79 s1.79\text{ s}

C

1.60 s1.60\text{ s}

D

2.24 s2.24\text{ s}

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