Practice Start Test

Simple Harmonic Motion

NEET Physics · 94 questions · Page 10 of 10 · Click an option or "Show Solution" to reveal answer

Q91
Motion of a particle in x-y plane is described by a set of following equations x=4sin⁡(π2−ωt) mx = 4\sin \left( {{\pi \over 2} - \omega t} \right)\,m and y=4sin⁡(ωt) my = 4\sin (\omega t)\,m. The path of the particle will be :
A circular
B helical
C parabolic
D elliptical
Correct Answer
Option A
Solution
x=4sin⁡(π2−ωt)x = 4\sin \left( {{\pi \over 2} - \omega t} \right)
=4cos⁡(ωt)= 4\cos (\omega t)
y=4sin⁡(ωt)y = 4\sin (\omega t)
⇒x2+y2=42\Rightarrow {x^2} + {y^2} = {4^2}

⇒\Rightarrow The particle is moving in a circular motion with radius of 4 m.

Q92
A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1/1000 of the original amplitude is close to :-
A 100 s
B 10 s
C 20 s
D 50 s
Correct Answer
Option C
Solution

Time for 10 oscillations =

105=2 s{{10} \over 5} = 2\,s

A = A0 e–kt

12=e−2k⇒ln⁡2=2k{1 \over 2} = {e^{ - 2k}} \Rightarrow \ln 2 = 2k

10–3 = e–kt ⇒\Rightarrow 3In10 = kt

t=3ln⁡10k=3ln⁡10ln⁡2×2t = {{3\ln 10} \over k} = {{3\ln 10} \over {\ln 2}} \times 2

=

6×2.30.69≈20 s6 \times {{2.3} \over {0.69}} \approx 20\,s
Q93
Time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift moves upwards with an acceleration g/2, the time period of pendulum will be :
A 3T\sqrt 3 T
B 23T\sqrt {{2 \over 3}} T
C T3{T \over {\sqrt 3 }}
D 32T\sqrt {{3 \over 2}} T
Correct Answer
Option B
Solution

When lift is stationary

T=2πLgT = 2\pi \sqrt {{L \over g}}

A pseudo force will act downwards when lift is moving upwards. ∴\therefore

geff=g+g2=3g2{g_{eff}} = g + {g \over 2} = {{3g} \over 2}

∴\therefore New time period

T′=2πLgeffT' = 2\pi \sqrt {{L \over {{g_{eff}}}}}
T′=2π2L3gT' = 2\pi \sqrt {{{2L} \over {3g}}}

∴\therefore

T′=23TT' = \sqrt {{2 \over 3}} T
Q94
If a simple harmonic motion is represented by d2xdt2+αx=0.{{{d^2}x} \over {d{t^2}}} + \alpha x = 0. its time period is
A 2πα{{2\pi } \over {\sqrt \alpha }}
B 2πα{{2\pi } \over \alpha }
C 2πα2\pi \sqrt \alpha
D 2πα2\pi \alpha
Correct Answer
Option A
Solution
d2xdt2=−αx=−ω2x{{{d^2}x} \over {d{t^2}}} = - \alpha x = - {\omega ^2}x
⇒ω=α\Rightarrow \omega = \sqrt \alpha
    \,\,\,\,

or

    \,\,\,\,
T=2πω=2παT = {{2\pi } \over \omega } = {{2\pi } \over {\sqrt \alpha }}
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