Physics Quiz Online for Class 11 on “Oscillations – Velocity and Acceleration in Simple Harmonic Motion”.
1. A particle is executing SHM and currently going towards the amplitude. If it is at A/2, what is the relation between the direction of velocity and acceleration?
a) Both vectors point towards the amplitude
b) Velocity is towards amplitude & acceleration is towards mean position
c) Velocity is towards mean position & acceleration is towards the amplitude
d) Both vectors point towards the mean position
Answer: b
Clarification: The force on a particle in SHM is always towards the mean position. The particle is currently going towards an extreme, thus velocity will be towards the amplitude.
2. In a SHM, for what value of w, will the magnitude of maximum acceleration be greater than the magnitude of maximum velocity? Here, w is angular frequency.
a) w > 1
b) w < 1
c) w = 0
d) Not possible for any value of w
Answer: a
Clarification: Let the velocity equation of SHM be: v = Awcos(wt),
acceleration will be given by: a = -Aw2sin(wt).
Maximum magnitude of vel = Aw
Maximum magnitude of acc = Aw2.
For, Aw2 > Aw
w > 1.
3. A particle of mass m starts from the mean position of a SHM, at t=0, and goes towards -A. If the angular frequency of SHM is w, find the force acting on it as a function of time.
a) mAw2sin(wt)
b) mAw2sin(wt+π)
c) -mAw2cos(wt+π)
d) -mAw2cos(wt)
Answer: a
Clarification: The displacement equation will be given by: x = -Asin(wt).
On taking its derivative, we get:
v = -Awcos(wt).
Further, we get: a = Aw2sin(wt).
Thus, force is given by: F(t) = mAw2sin(wt).
4. The graph of acceleration vs time for a SHM is given. FInd the equation of position of the particle as a function of time. Assume that the particle is oscillating on the x-axis about the origin. And consider RHS of origin to be positive & LHS to be negative.
a) 8cos(πt)
b) -8cos(πt)
c) 0.81sin(πt)
d) -0.81sin(πt)
Answer: d
Clarification: The particle has zero acceleration at t=0, so it starts from the mean position. Then the acceleration becomes positive which implies that the particle is moving towards -A ( negative extreme). The equation of acceleration can be given by: a = 8sinwt.
The time period is 2s,
so w = 2π/2 = πs-1.
∴ a = 8sin(πt).
∴ v = -8/π cos(πt)
∴ x = -8/pi2sin(πt) = -0.81sin(πt).
5. A particle is undergoing a SHM of amplitude 10cm. What should be the minimum value of acceleration at an extreme position for maximum speed at centre to be 5m/s?
a) 20m/s2
b) 5m/s2
c) 0
d) 250m/s2
Answer: d
Clarification: vmax = Aw.
5 = 0.1w.
w = 50s-1.
amin at extreme = Aw2
= 0.1*50*50
= 250m/s2.
6. A particle in uniform circular motion is projected on its diameter. The motion of projection will be simple harmonic. Select the correct option regarding speed and acceleration of the particle in circular motion.
a) Speed is constant, acceleration is zero
b) Speed is constant, acceleration is non-zero
c) Speed changes, acceleration is non-zero
d) Speed changes, acceleration is zero
Answer: a
Clarification: For the particle in uniform circular motion, magnitude of velocity is constant, but direction is continuously changing. Thus, speed is constant but acceleration is non-zero.
7. The curves given below are that of position(x), velocity(v) & acceleration(a) vs time. What can we infer about the value of angular frequency ‘f’ from the given curves?
a) f > 1
b) f < 1
c) f = 0
d) We need the time period in order to find ‘f’
Answer: a
Clarification: Amplitude of position = A
Amplitude of velocity = Af
Amplitude of acceleration = Af2.
From the graph we deduce that the amplitude of a is greater than the amplitude of v which in turn is greater than the amplitude of x.
Thus, angular frequency will be greater than 1.
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