Q#11
The total mechanical energy of a spring-mass system in simple harmonic motion is E = ½m⍵²A². Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude A remains the same. The new mechanical energy will(a) become 2E
(b) become E/2
(c) become √2E
(d) remain E.
Answer: (d)
Since no external work is done on the system the energy of the system will remain the same. Hence option (d).
Q#12
The average energy in one time period in simple harmonic motion is
(a) ½m⍵²A²
(b) ¼m⍵²A²
(c) m⍵²A²
(d) zero
Answer: (a)
The total energy at any instant in a simple harmonic motion is constant and equal to ½m⍵²A². Hence the average energy in one time period will also be the same = ½m⍵²A². So, option (a).
Q#13
A particle executes simple harmonic motion with a frequency ν. The frequency with which the kinetic energy oscillates is
(a) ν/2
(b) ν
(c) 2ν
(d) zero
Answer: (c)
Between one extreme position to another i.e. in half the time period the KE increases from zero to maximum (at the mean position) and again to zero thus completing one cycle. Therefore the time period of oscillation of the KE = T/2 where T is the time period of the SHM. Since the frequency of SHM ν = 1/T, the frequency of oscillation of the K.E. ν'= 1/(T/2)= 2(1/T) = 2ν. Hence the option (c).
Q#14
A particle executes simple harmonic motion under the restoring force provided by a spring. The time period is T. If the spring is divided into two equal parts and one part is used to continue the simple harmonic motion, the time period will
(a) remain T
(b) become 2T
(c) become T/2
(d) become T/√2
Answer: (d)
The time period of a spring-mass system is given as
T = 2π√(m/k)
When the spring is divided in two equal halves the spring constant of each half doubles, so k'=2k.
New time period
T' = 2π√{m/(2k)} = (1/√2){2π√(m/k)}
T'= T/√2
Hence the option (d).
Q#15
Two bodies A and B of equal mass are suspended from two separate massless springs of spring constant k₁ and k₂ respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that of B is
(a) k₁/k₂
(b) √(k₁/k₂)
(c) k₂/k₁
(d) √(k₂/k₁)
Answer: (d)
The maximum velocity occurs at the mean position where the total energy is in the form of kinetic energy = ½mv². Since the maximum velocities of both systems are equal, their total energy is also equal.
At the extreme position, the total energy is in the form of potential energy. These will also be equal. Let amplitude of A = A' and that of B = B'. Equating the total energy at extremes we get,
½k₁A'² = ½k₂B'²
A'²/B'² = k₂/k₁
A'/B' = √(k₂/k₁)
Hence the option (d).
Q#16
A spring-mass system oscillates with a frequency ν. If it is taken in an elevator slowly accelerating upward, the frequency will
(a) increase
(b) decrease
(c) remain same
(d) become zero.
Answer: (c)
The time period of a spring-mass system is given by
T = 2π√(m/k)
Since this expression is not dependent on the acceleration due to the gravity, the time period T and hence the frequency (since ν = 1/T) remains the same. The option (c) is correct.
Q#17
A spring-mass system oscillates in a car. If the car accelerates on a horizontal road, the frequency of oscillation will
(a) increase
(b) decrease
(c) remain same
(d) become zero.
Answer: (c)
The frequency of a spring-mass system depends only on the mass and the spring constant, none of which changes in the accelerating car. So, the frequency remains the same. Option (c).
Q#18
A pendulum clock that keeps correct time on the earth is taken to the moon. It will run
(a) at correct rate
(b) 6 times faster
(c) √6 times faster
(d) √6 times slower.
Answer: (d)
The time period of the pendulum of the clock on the earth T =2π√(l/g). On the moon g'=g/6. The time period on the moon
T' = 2π√6l/g) =√6{2π√(l/g)} = √6T
So the pendulum will take √6 times more time in one oscillation on the moon and it will run √6 times slower. The option (d) is true.
Q#19
A wall clock uses a vertical spring-mass system to measure the time. Each time the mass reaches an extreme position, the clock advances by a second. The clock gives the correct time at the equator. If the clock is taken to the poles it will
(a) run slow
(b) run fast
(c) stop working
(d) give correct time.
Answer: (d)
Same explanation as in Q-17. Option (d) is true.
Q#20
A pendulum clock keeping correct time is taken to high altitudes,
(a) it will keep correct time
(b) its length should be increased to keep correct time
(c) its length should be decreased to keep correct time
(d) it cannot keep correct time even if the length is changed.
Answer: (c)
At high altitudes, the value of g decreases. Since the time period of the pendulum T = 2π√(l/g), it will increase at high altitudes. The clock will not give the correct time. To keep the correct time T should not change and to keep T constant the ratio l/g should not change. Since the value of g is getting decreased at the high altitudes, the length l should also be proportionately decreased to keep the ratio l/g constant. Hence the option (c) is true.
Q#21
The free end of a simple pendulum is attached to the ceiling of a box. The box is taken to a height and the pendulum is oscillated. When the bob is at its lowest point, the box is released to fall freely. As seen from the box during this period, the bob will
(a) continue its oscillation as before
(b) stop
(c) will go in a circular path
(d) move on a straight line.
Answer: (c)
At the lowest point the bob is moving horizontally with the maximum speed. When we observe the bob from the box when free falling, the box is a non-inertial frame with acceleration g downwards. If we want to apply newtons laws of motion here we will have to apply a pseudo force mg upwards. Thus the weight mg and the pseudo force -mg have a resultant zero force on the bob and hence no restoring force for the oscillation. Since the bob has a horizontal speed and it is tied to the ceiling with a string it will go in a circular path. Hence the option (c) is true.
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