The velocity of a particle moving with simple harmonic motion is____ at the mean position.
V = ω√(A2 – x2)
So the velocity is maximum at mean position
The periodic time (tp) is given by
Periodic time is the time taken for one complete revolution of the particle.
∴ Periodic time, tp = 2 π/ω seconds.
A frequency of 1Hz corresponds to:
Frequency is defined as time taken to perform one oscillation by the object, hence 1Hz corresponds to 1 vibration per sec.
A second pendulum is mounted in a space shuttle. Its period of oscillations will decrease when rocket is
Time Period, T = 2π ; where l = Length of seconds pendulum
g’= Apparent Gravity
=> For the period of oscillations of Seconds Pendulum to decrease, the Apparent gravity (g’) has to increase because-
Hence, Time Period of oscillations of Seconds Pendulum will decrease when the rocket is ascending up with uniform acceleration.
A particle of mass 10 gm lies in a potential field v = 50 x2 + 100. The value of frequency of oscillations in Hz is
If a simple pendulum oscillates with an amplitude 50 mm and time period 2s, then its maximum velocity is
We know that in a simple harmonic motion the maximum velocity, vmax = A⍵
Here A = 50 mm
And ⍵ = 2π / T
= 2π / 2
Hence vmax = 50 x 10-3.π
= 0.15 m/s
In simple harmonic motion the displacement of a particle from its equilibrium position is given by . Here the phase of motion is
Find the amplitude of the S.H.M whose displacement y in cm is given by equation y= 3sin 157t +4cos157t where t is time in seconds.
When the displacement of a SHM is y=a sin wt+ b cos wt, the amplitude of the SHM
will be A=√a2+b2
Here, a=3, b=4 so amplitude A=√32+42=5 cm
Hence option B is correct.
What will be the phase difference between bigger pendulum (with time period 5T/4 )and smaller pendulum (with time period T) after one oscillation of bigger pendulum?
After one oscillation of a bigger pendulum i.e. 5T/4, ¼ of the total phase is travelled by the smaller pendulum while the bigger is still at initial position. Thus the phase difference between two is ¼ (2π) - 0 = π/2
A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:
The magnitudes of the velocity and acceleration of the particle when its displacement is 'y' are and ω2y respectively.