Damped Simple Harmonic Motion
PHXI14:OSCILLATIONS

364275 If the maximum velocity and maximum acceleration of a particle executing SHM are equal in magnitude, the time period will be :-

1 \(12.56\,{\rm{sec}}\)
2 \(1.57\,{\rm{sec}}\)
3 \(6.28\,{\rm{sec}}\)
4 \(3.14\,{\rm{sec}}\)
PHXI14:OSCILLATIONS

364276 A S.H.M. has amplitude ' \(a\) ' and time period \(T\). The maximum velocity will be

1 \(\dfrac{4 a}{T}\)
2 \(\dfrac{2 a}{T}\)
3 \(2 \pi \sqrt{\dfrac{a}{T}}\)
4 \(\dfrac{2 \pi a}{T}\)
PHXI14:OSCILLATIONS

364277 Match the Column I (quantity) with Column II (value) for an object executing simple harmonic motion in a horizontal plane with displacement given as \(x=A \cos \omega t\) and select the correct answer from the codes given below
Column I
Column II
A
\(\dfrac{v_{\max }}{A}\)
P
\(T / 8\)
B
\(\dfrac{a_{\max }}{A}\)
Q
\(T / 12\)
C
\(\begin{array}{l}\text { If object starts from } \\x=+A, \text { then time to } \\\text { reach at } \dfrac{A}{\sqrt{2}}\end{array}\)
R
\(\omega\)
D
\(\begin{array}{l}\text { If object starts from } \\x=0 \text { and move towards } \\\text { right, then the time to } \\\text { reach at }+A / 2\end{array}\)
S
\(\omega^{2}\)

1 A - R, B - P, C - S, D - Q
2 A - R, B - S, C - P, D - Q
3 A - S, B - R, C - Q, D - P
4 A - R, B - S, C - Q, D - P
PHXI14:OSCILLATIONS

364278 The acceleration \(a\) of a particle undergoing S.H.M. is shown in the figure. Which of the labelled points corresponds to the particle being at \( - {x_{\max }}\)
supporting img

1 3
2 4
3 1
4 2
PHXI14:OSCILLATIONS

364275 If the maximum velocity and maximum acceleration of a particle executing SHM are equal in magnitude, the time period will be :-

1 \(12.56\,{\rm{sec}}\)
2 \(1.57\,{\rm{sec}}\)
3 \(6.28\,{\rm{sec}}\)
4 \(3.14\,{\rm{sec}}\)
PHXI14:OSCILLATIONS

364276 A S.H.M. has amplitude ' \(a\) ' and time period \(T\). The maximum velocity will be

1 \(\dfrac{4 a}{T}\)
2 \(\dfrac{2 a}{T}\)
3 \(2 \pi \sqrt{\dfrac{a}{T}}\)
4 \(\dfrac{2 \pi a}{T}\)
PHXI14:OSCILLATIONS

364277 Match the Column I (quantity) with Column II (value) for an object executing simple harmonic motion in a horizontal plane with displacement given as \(x=A \cos \omega t\) and select the correct answer from the codes given below
Column I
Column II
A
\(\dfrac{v_{\max }}{A}\)
P
\(T / 8\)
B
\(\dfrac{a_{\max }}{A}\)
Q
\(T / 12\)
C
\(\begin{array}{l}\text { If object starts from } \\x=+A, \text { then time to } \\\text { reach at } \dfrac{A}{\sqrt{2}}\end{array}\)
R
\(\omega\)
D
\(\begin{array}{l}\text { If object starts from } \\x=0 \text { and move towards } \\\text { right, then the time to } \\\text { reach at }+A / 2\end{array}\)
S
\(\omega^{2}\)

1 A - R, B - P, C - S, D - Q
2 A - R, B - S, C - P, D - Q
3 A - S, B - R, C - Q, D - P
4 A - R, B - S, C - Q, D - P
PHXI14:OSCILLATIONS

364278 The acceleration \(a\) of a particle undergoing S.H.M. is shown in the figure. Which of the labelled points corresponds to the particle being at \( - {x_{\max }}\)
supporting img

1 3
2 4
3 1
4 2
PHXI14:OSCILLATIONS

364275 If the maximum velocity and maximum acceleration of a particle executing SHM are equal in magnitude, the time period will be :-

1 \(12.56\,{\rm{sec}}\)
2 \(1.57\,{\rm{sec}}\)
3 \(6.28\,{\rm{sec}}\)
4 \(3.14\,{\rm{sec}}\)
PHXI14:OSCILLATIONS

364276 A S.H.M. has amplitude ' \(a\) ' and time period \(T\). The maximum velocity will be

1 \(\dfrac{4 a}{T}\)
2 \(\dfrac{2 a}{T}\)
3 \(2 \pi \sqrt{\dfrac{a}{T}}\)
4 \(\dfrac{2 \pi a}{T}\)
PHXI14:OSCILLATIONS

364277 Match the Column I (quantity) with Column II (value) for an object executing simple harmonic motion in a horizontal plane with displacement given as \(x=A \cos \omega t\) and select the correct answer from the codes given below
Column I
Column II
A
\(\dfrac{v_{\max }}{A}\)
P
\(T / 8\)
B
\(\dfrac{a_{\max }}{A}\)
Q
\(T / 12\)
C
\(\begin{array}{l}\text { If object starts from } \\x=+A, \text { then time to } \\\text { reach at } \dfrac{A}{\sqrt{2}}\end{array}\)
R
\(\omega\)
D
\(\begin{array}{l}\text { If object starts from } \\x=0 \text { and move towards } \\\text { right, then the time to } \\\text { reach at }+A / 2\end{array}\)
S
\(\omega^{2}\)

1 A - R, B - P, C - S, D - Q
2 A - R, B - S, C - P, D - Q
3 A - S, B - R, C - Q, D - P
4 A - R, B - S, C - Q, D - P
PHXI14:OSCILLATIONS

364278 The acceleration \(a\) of a particle undergoing S.H.M. is shown in the figure. Which of the labelled points corresponds to the particle being at \( - {x_{\max }}\)
supporting img

1 3
2 4
3 1
4 2
PHXI14:OSCILLATIONS

364275 If the maximum velocity and maximum acceleration of a particle executing SHM are equal in magnitude, the time period will be :-

1 \(12.56\,{\rm{sec}}\)
2 \(1.57\,{\rm{sec}}\)
3 \(6.28\,{\rm{sec}}\)
4 \(3.14\,{\rm{sec}}\)
PHXI14:OSCILLATIONS

364276 A S.H.M. has amplitude ' \(a\) ' and time period \(T\). The maximum velocity will be

1 \(\dfrac{4 a}{T}\)
2 \(\dfrac{2 a}{T}\)
3 \(2 \pi \sqrt{\dfrac{a}{T}}\)
4 \(\dfrac{2 \pi a}{T}\)
PHXI14:OSCILLATIONS

364277 Match the Column I (quantity) with Column II (value) for an object executing simple harmonic motion in a horizontal plane with displacement given as \(x=A \cos \omega t\) and select the correct answer from the codes given below
Column I
Column II
A
\(\dfrac{v_{\max }}{A}\)
P
\(T / 8\)
B
\(\dfrac{a_{\max }}{A}\)
Q
\(T / 12\)
C
\(\begin{array}{l}\text { If object starts from } \\x=+A, \text { then time to } \\\text { reach at } \dfrac{A}{\sqrt{2}}\end{array}\)
R
\(\omega\)
D
\(\begin{array}{l}\text { If object starts from } \\x=0 \text { and move towards } \\\text { right, then the time to } \\\text { reach at }+A / 2\end{array}\)
S
\(\omega^{2}\)

1 A - R, B - P, C - S, D - Q
2 A - R, B - S, C - P, D - Q
3 A - S, B - R, C - Q, D - P
4 A - R, B - S, C - Q, D - P
PHXI14:OSCILLATIONS

364278 The acceleration \(a\) of a particle undergoing S.H.M. is shown in the figure. Which of the labelled points corresponds to the particle being at \( - {x_{\max }}\)
supporting img

1 3
2 4
3 1
4 2