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Engineering
264 Views   |

If the length of a simple pendulum is doubled keeping its amplitude constant its energy will be

• Option 1)

Unchanged

• Option 2)

Doubled

• Option 3)

Four times

• Option 4)

Halved

As we learnt in Time period of oscillation of simple pendulum - - wherein l = length of pendulum  g = acceleration due to gravity.    Halved Option 1) Unchanged This solution is incorrect. Option 2) Doubled This solution is incorrect. Option 3) Four times This solution is incorrect. Option 4) Halved This solution is correct.
Engineering
357 Views   |

Two springs of constants k1 and k2 equal maximum velocities, when executing simple harmonic motion. The ratio of their amplitudes (masses are equal) will be

• Option 1)

$\frac{k_1}{k_2}$

• Option 2)

$\frac{k_2}{k_1}$

• Option 3)

$(\frac{k_1}{k_2})^{\frac{1}{2}}$

• Option 4)

$(\frac{k_2}{k_1})^{\frac{1}{2}}$

As learnt in Relation of velocity and displacement -   - wherein   x is displacement from mean position   A is Amplitude.                                                              Option 1) This option is incorrect Option 2) This option is incorrect Option 3) This option is incorrect Option 4) This option is correct
Engineering
1265 Views   |

Two simple pendulums of length 5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed how many oscillations

• Option 1)

5

• Option 2)

2

• Option 3)

1

• Option 4)

3

As learnt in Time period of oscillation of simple pendulum - - wherein l = length of pendulum  g = acceleration due to gravity.     Option 1) 5 This option is incorrect Option 2) 2 This option is correct Option 3) 1 This option is incorrect Option 4) 3 This option is incorrect
Engineering
123 Views   |

Two bodies M and N of equal masses are suspended from two separate mass less springs of spring constants $k_1$ and $k_2$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitudes of M to that of N is

• Option 1)

$\frac{k_1}{k_2}$

• Option 2)

$\sqrt{\frac{k_1}{k_2}}$

• Option 3)

$\frac{k_2}{k_1}$

• Option 4)

$\sqrt{\frac{k_2}{k_1}}$

As learnt in Relation of velocity and displacement -   - wherein   x is displacement from mean position   A is Amplitude.                                                      Option 1) This option is incorrect Option 2) This option is incorrect Option 3) This option is incorrect Option 4) This option is correct
Engineering
84 Views   |

The period of thin magnet is 4 sec. if it is divided into two equal halves then the time period of each part will be

• Option 1)

4 sec

• Option 2)

1 sec

• Option 3)

2 sec

• Option 4)

8 sec

As we discussed in concept Physical pendulum - Any rigid body suspended from a fixed support constitute a physical pendulum. - wherein e.g. A circular ring suspended on a nail in a wall etc.    if it is divided into two equal halves the time period will be half T = 2sec Option 1) 4 sec Option is incorrect Option 2) 1 sec Option is incorrect Option 3) 2 sec Option is correct Option 4) 8...
Engineering
85 Views   |

The maximum velocity of a body in S.H.M. is 0.25 m/s and maximum acceleration is 0.75 $ms^{-2}$, the period of S.H.M. is

• Option 1)

$\frac{\pi}{3}\, \, second$

• Option 2)

$\frac{\pi}{2}\, \, second$

• Option 3)

$\frac{2\pi}{3}\, \, second$

• Option 4)

$\pi\, \, second$

As we discussed in concept Relation of velocity and displacement -   - wherein   x is displacement from mean position   A is Amplitude.     Option 1) Option is incorrect Option 2) Option is incorrect Option 3) Option is correct Option 4) Option is incorrect
Engineering
88 Views   |

The maximum velocity and maximum acceleration of a body moving in a simple harmonic oscillator are 2m/s and 4 $ms^{-2}$ the angular velocity is

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we discussed in concept Relation of velocity and displacement -   - wherein   x is displacement from mean position   A is Amplitude.       Option 1) 1 rad/s Option is incorrect Option 2) 2 rad/s Option is correct Option 3) 4 rad/s Option is incorrect Option 4) 5 rad/s Option is incorrect
Engineering
314 Views   |

The angular frequency of small oscillations of the system shown in the figure is

• Option 1)

√(K / 2m)

• Option 2)

√(2K / m)

• Option 3)

√(K / 4m)

• Option 4)

√(4K / m)

Option 1) √(K / 2m) Option is incorrect Option 2) √(2K / m) Option is incorrect Option 3) √(K / 4m) Option is correct Option 4) √(4K / m) Option is incorrect
Engineering
130 Views   |

The acceleration due to gravity changes from 9.8 $ms^{-2}$ to 9.5 $ms^{-2}$. To keep the period of pendulum constant, its length must changes by

• Option 1)

3 m

• Option 2)

0.3 m

• Option 3)

0.3 cm

• Option 4)

3 cm

As discussed in Time period of oscillation of simple pendulum - - wherein l = length of pendulum  g = acceleration due to gravity.    g changes from  to  length changes by 3 cm Option 1) 3 m This solution is incorrect. Option 2) 0.3 m This solution is incorrect. Option 3) 0.3 cm This solution is incorrect. Option 4) 3 cm This solution is correct.
Engineering
169 Views   |

Starting from the extreme position, the time taken by an ideal simple pendulum to travel a distance of half of the amplitude is

• Option 1)

T/6

• Option 2)

T/12

• Option 3)

T/13

• Option 4)

T/4

As discussed in Phase - The quantity is called the phase . It determines the status of the particle in simple harmonic motion.     - wherein e.g.  phase        Option 1) T/6 This solution is correct. Option 2) T/12 This solution is incorrect. Option 3) T/13 This solution is incorrect. Option 4) T/4 This solution is incorrect.
Engineering
838 Views   |

If the length of simple pendulum is increased by 44% then what is the change in the time period of the pendulum?

• Option 1)

22 %

• Option 2)

20 %

• Option 3)

33 %

• Option 4)

44 %

As discussed in Time period of oscillation of simple pendulum - - wherein l = length of pendulum  g = acceleration due to gravity.    length increased by 44%.  New length  i.e. Time period increased by 0.2 or 20% Option 1) 22 % This solution is incorrect. Option 2) 20 % This solution is correct. Option 3) 33 % This solution is incorrect. Option 4) 44 % This solution is incorrect.
Engineering
295 Views   |

If a simple pendulum oscillates with an amplitude of 50 mm and time period of 2s, then its maximum velocity is

• Option 1)

0.10 m/s

• Option 2)

0.16 m/s

• Option 3)

0.25 m/s

• Option 4)

0.50 m/s

As discussed in Relation of velocity and displacement -   - wherein   x is displacement from mean position   A is Amplitude.       Option 1) 0.10 m/s This solution is incorrect. Option 2) 0.16 m/s This solution is correct. Option 3) 0.25 m/s This solution is incorrect. Option 4) 0.50 m/s This solution is incorrect.
Engineering
108 Views   |

If a simple harmonic oscillator has got a displacement of 0.02 m and acceleration equal to 2.0 $ms^{-2}$ at any time, the angular frequency of the oscillator is equal to

• Option 1)

10 rad $s^{-1}$

• Option 2)

0.1 rad $s^{-1}$

• Option 3)

100 rad $s^{-1}$

• Option 4)

1 rad $s^{-1}$

As we discussed in concept Equation of S.H.M. -   - wherein                                                        => Option 1) 10 rad  This option is correct. Option 2) 0.1 rad  This option is incorrect. Option 3) 100 rad  This option is incorrect. Option 4) 1 rad  This option is incorrect.
Engineering
160 Views   |

If a bar magnet of magnetic moment M is kept in a uniform magnetic field B, its time period of oscillation is T. Another magnet of same length and breadth is kept in a same magnetic field. If magnetic moment of new magnet is M/4, then its oscillation time period is

• Option 1)

T

• Option 2)

2T

• Option 3)

T/2

• Option 4)

T/4

As we discussed in concept Time period of physical pendulum - - wherein moment of inertia about point of hinge Seperation between point of suspension & centre of mass                                                    Option 1) T This option is incorrect. Option 2) 2T This option is correct. Option 3) T/2 This option is incorrect. Option 4) T/4 This option is incorrect.
Engineering
208 Views   |

For a particle executing simple harmonic motion, the kinetic energy K is given by $K=K_0\cos ^{2}wt$ the maximum value of potential energy is

• Option 1)

$K_0$

• Option 2)

Zero

• Option 3)

$\frac{K_0}{2}$

• Option 4)

Not obtainable

As we discussed in concept Average value of kinetic energy - Average of kinetic energy     - wherein                                                                  Option 1) This option is correct. Option 2) Zero This option is incorrect. Option 3) This option is incorrect. Option 4) Not obtainable This option is incorrect.
Engineering
173 Views   |

A spring has a force constant K and a mass m is suspended from it. The spring is cut into half and the same a mss is suspended from one of the halves. If the frequency of oscillation in the first case is a, then the frequency in the second case will be

• Option 1)

2 a

• Option 2)

a

• Option 3)

a/2

• Option 4)

$a\sqrt{2}$

As we discussed in concept Time period of oscillation for spring mass system - - wherein m = mass of block K = spring constant     when spring cut into two half then the force constant will be doubled. Option 1) 2 a     This option is incorrect. Option 2) a This option is incorrect. Option 3) a/2 This option is incorrect. Option 4) This option is correct.
Engineering
125 Views   |

A simple harmonic oscillator has an amplitude a and time period T. The time required to travel from $x= \frac{a}{2}$  is

• Option 1)

$\frac{T}{6}$

• Option 2)

$\frac{T}{2}$

• Option 3)

$\frac{T}{4}$

• Option 4)

$\frac{T}{3}$

As we discussed in concept Phase - The quantity is called the phase . It determines the status of the particle in simple harmonic motion.     - wherein e.g.  phase     Option 1) Option is Correct Option 2) Option is incorrect Option 3) Option is incorrect Option 4) Option is incorrect
Engineering
94 Views   |

A simple harmonic motion having an amplitude A and time period T is represented by the equation y = 5 sin p (t + 4)m then the values of (A in m) and (T in  sec) are

• Option 1)

A = 5, T = 2

• Option 2)

A = 10, T = 1

• Option 3)

A = 5, T = 1

• Option 4)

A = 10, T = 2

As we discussed in concept Phase - The quantity is called the phase . It determines the status of the particle in simple harmonic motion.     - wherein e.g.  phase       Option 1) A = 5, T = 2 Option is correct Option 2) A = 10, T = 1 Option is incorrect Option 3) A = 5, T = 1 Option is incorrect Option 4) A = 10, T = 2 Option is incorrect
Engineering
173 Views   |

A particle of mass m is executing SHM about its mean position. The total energy of the particle at given instant is

• Option 1)

$\frac{\pi^{2}mA^{2}}{T^{2}}$

• Option 2)

$\frac{4\pi^{2}mA^{2}}{T^{2}}$

• Option 3)

$\frac{2\pi^{2}mA^{2}}{T^{2}}$

• Option 4)

$\frac{8\pi^{2}mA^{2}}{T^{2}}$

As we discussed in concept Total energy in S.H.M. - Total Energy = Kinetic + Potential Energy - wherein Total Energy = Hence total energy in S.H.M. is constant    Total energy =  Option 1) Option is incorrect Option 2) Option is incorrect Option 3) Option is correct Option 4) Option is incorrect
Engineering
126 Views   |

A particle executes simple pendulum harmonic motion of amplitude A. at what distance from the mean position is its kinetic energy to its potential energy?

• Option 1)

0.51 A

• Option 2)

0.61 A

• Option 3)

0.71 A

• Option 4)

0.81 A

As we discussed in concept Kinetic energy in S.H.M. -   - wherein    K=U   Option 1) 0.51 A Option is incorrect Option 2) 0.61 A Option is incorrect Option 3) 0.71 A Option is correct Option 4) 0.81 A Option is incorrect
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