Q#1
A student says that he had applied a force F = -k/x on a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he has worked only with positive x and no other force acted on the particle.(a) As x increases k increases
(b) As x increases k decreases
(c) As x increases k remains constant.
(d) The motion cannot be simple harmonic.
Answer: (a)
Since the motion is simple harmonic, the force F = -⍵²x where ⍵² is a constant. Equating the force applied by the student we get,
-k/x = -⍵²x
k = ⍵²x²
It is clear from this relation that if x increases k increases.
Q#2
The time period of a particle in simple harmonic motion is equal to the time between consecutive appearances of the particle at a particular point in its motion. Thi point is
(a) the mean position
(b) an extreme position
(c) between the mean position and the positive extreme
(d) between the mean position and the negative extreme.
Answer: (b)
At the extreme position the particle is in the same phase at consecutive appearance, hence (b) is true.
In all other options, the consecutive appearance of the particle is not in the same phase. All other options are not true.
Q#3
The time period of a particle in simple harmonic motion is equal to the smallest time between the particle acquiring a particular velocity v. The value of v is
(a) vₘₐₓ
(b) 0
(c) between 0 and vₘₐₓ
(d) between 0 and -vₘₐₓ
Answer: (a)
Velocity is a vector, means it has magnitude as well as direction. The maximum velocity at the mean position is acquired by the particle after a single time period at the same position. Other velocities occur twice between one extreme to other extreme position (i.e. within half a time period, See picture below). Hence only (a) true, others not true.
Q#4
The displacement of a particle in simple harmonic motion in one time period is
(a) A
(b) 2A
(c) 4A
(d)zero
The displacement of a particle in simple harmonic motion in one time period is
(a) A
(b) 2A
(c) 4A
(d)zero
Answer: (d)
The displacement is a vector and its magnitude is given by the line joining the initial position to the final position. Since after one time period the particle is at its initial position, hence it is zero. Option(d).
Q#5
The distance moved by a particle in simple harmonic motion in one time period is
(a) A
(b) 2A
(c) 4A
(d)zero
Answer: (c)
The distance moved is a scalar quantity and in a simple harmonic motion the distance covered in one time period is 4A.
For example, if the particle is at the left extreme at t = 0 then at t = T it will again come back to that point.
The distance covered = 2A + 2A = 4A.
Q#6
The average acceleration in one time period in a simple harmonic motion is
(a) A⍵²
(b) A⍵²/2
(c) A⍵²/√2
(d) zero
Answer: (d)
Since the acceleration is proportional to the displacement and always directed towards the mean position, half of one time period it is directed towards left and in rest half, it is towards the right. Hence the average acceleration in one time period is zero. Option (d).
Q#7
The motion of a particle is given by x = Asin⍵t + Bcos⍵t. The motion of the particle is
(a) not simple harmonic
(b) simple harmonic with amplitude A + B
(c) simple harmonic with amplitude (A + B)/2
(d) simple harmonic with amplitude √(A² + B²).
Answer: (d)
The given equation can be written as,
x = √(A² + B²){(A/√(A² + B²))sin⍵t+(B/√(A² + B²))cos⍵t}
Since the magnitudes of A/√(A² + B²) and B/√(A² + B²) are less than 1 and the sum of their square is also 1 we can find an angle α between 0 and 2π such that
sinα = B/√(A² + B²) and cosα = A/√(A² + B²)
So, x = √(A² + B²){sin⍵t(cosα) + cos⍵t(sinα)}
x = √(A² + B²)sin(⍵t + α)
This is an equation of simple harmonic motion with amplitude √(A² + B²). Hence option (d).
Q#8
The displacement of a particle is given by r = A(icos⍵t + jsin⍵t). The motion of the particle is
(a) simple harmonic
(b) on a straight line
(c) on a circle
(d) with constant acceleration.
Answer: (c)
The magnitude of the displacement
r = √(A²cos²⍵t + A²sin²⍵t) = √A² = A
The direction of the displacement from X-axis is
tanα = Asin⍵t/Acos⍵t = tan⍵t
α = ⍵t
It means the position vector of the particle can take any angle but its magnitude is constant. So the motion is on a circle with center at the origin. Option (c).
A particle moves on the X-axis according to the equation x = A + Bsin⍵t. The motion is simple harmonic with amplitude
(a) A
(b) B
(c) A + B
(d) √(A² + B²)
Answer: (b)
The equation can be written as
x – A = Bsin⍵t
If x – A = y then
y = Bsin⍵t
This is the equation of a simple harmonic motion with amplitude B. Hence option(b).
Figure (12-Q1) represents two simple harmonic motions.
The parameter which has different values in the two motions is
(a) amplitude
(b) frequency
(c) phase
(d) maximum velocity.
Answer: (c)
It is clear from the figure that in the next step when one time period is complete the particles will be back to their original position. Hence their frequency is the same. The maximum displacement from the mean position for both is the same, hence the amplitude is also the same. For the same amplitude and frequency, the maximum velocity is also the same. So it is the phase which is different. At the mean position, both particles are at the same time but in opposite direction. Hence their phase differs by an angle π. So option(c).
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