# SHM:The Calculus Picture

## Trending Questions

**Q.**A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6 s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is

**Q.**A body executing simple harmonic motion has a maximum acceleration equal to 24 metres/sec2 and maximum velocity equal to 16 metres/sec2. The amplitude of the simple harmonic motion is

**Q.**

If a simple harmonic motion is represented by d2xdt2+αx=0 its time period is

2πα

2π√α

2πα

2π√α

**Q.**Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion?

- When v is maximum, a is maximum
- Value of a is zero, whatever may be the value of v
- When v is maximum, a is zero
- When v is zero, a is zero

**Q.**The time period of simple pendulum is T. If its length is increased by 2 the new time period becomes

- 0⋅99 T
- 1⋅01 T
- 1⋅02 T
- 0⋅98 T

**Q.**

If the displacement x and velocity v of a particle executing simple harmonic motion are related through the expression 4v2=25−x2, then its time period is

π

6π

4π

2π

**Q.**The equation of a particle executing SHM is given by x=150sin(π3t+π2), where the amplitude is in metres and angular frequency is in rad/s. Find the maximum velocity (vmax) and maximum acceleration (amax) attained by the particle.

- None of the above
- vmax=50π m/s, amax=50π23 m/s2
- vmax=100π m/s, amax=50π23 m/s2
- vmax=50π m/s, amax=100π23 m/s2

**Q.**Two simple harmonic motions are given by x1=a2sinωt+√3a2cosωt and x2=asinωt+acosωt. What is the ratio of the amplitude of the first motion to that of the second ?

- √32
- 1√2
- √2
- 12

**Q.**Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion?

- When v is maximum, a is maximum
- Value of a is zero, whatever may be the value of v
- When v is zero, a is zero
- When v is maximum, a is zero

**Q.**The equation of a particle executing SHM is given by x=150sin(π3t+π2), where the amplitude is in metres and angular frequency is in rad/s. Find the maximum velocity (vmax) and maximum acceleration (amax) attained by the particle.

- vmax=50π m/s, amax=50π23 m/s2
- vmax=100π m/s, amax=50π23 m/s2
- vmax=50π m/s, amax=100π23 m/s2
- None of the above

**Q.**The time period of the simple harmonic motion represented by the equation d2xdt2+αx=0 is

- 2
- 2
- 2
- 2

**Q.**Find the velocity when KE=PE of a body undergoing SHM. Given, amplitude =x0 and angular frequency is ω. Also, how many times in a cycle is KE=PE ?

- ωx0√2, 2
- ωx0, 2
- ωx0√2, 4
- ωx0, 4

**Q.**

Given a differential equation of SHM :

ad2xdt2+bx=0

- 2π√b
- 2π√b
- 2π√b/a
- 2π√a/b

**Q.**If a simple harmonic motion is represented by d2xdt2+αx=0. Its time period is

- 2πα
- 2π√α
- 2πα
- 2π√α

**Q.**The time period of the simple harmonic motion represented by the equation d2xdt2+αx=0 is

- 2 πα
- 2 π√α
- 2 π/α
- 2 π/√α

**Q.**From the differential equation, d2xdt2+16x=0

(1) we can find time period of oscillation

(2) we cannot find time period of oscillation

(3) we cannot find initial phase of oscillation

(4) we cannot find frequency of oscillation.

- 1 & 2
- 2 & 3
- 1 & 3
- 1, 2 and 3

**Q.**A string oscillates according to the equation y′=(0.50cm)sin[(π3cm−1)x]cos[(40πs−1)t]. What is the amplitude?

**Q.**

If a simple harmonic motion is represented by d2xdt2+αx=0 its time period is

2πα

2π√α

2πα

2π√α

**Q.**

If the displacement x and velocity v of a particle executing simple harmonic motion are related through the expression 4v2=25−x2, then its time period is

π

2π

4π

6π

**Q.**The time taken by a particle executing simple harmonic motion to move from the mean position to (1/4)th of the maximum displacement is 1 s. Calculate the angular frequency of oscillation.

[sin−1(1/4)=0.252]

- 0.25 rads
- 0.32 rads
- 0.15 rads
- 0.20 rads

**Q.**The displacement x (in metres) of a particle is simple harmonic motion is related to time t (in seconds) as x = 0.01 cos(πt+π4) What is the frequency of the motion?

- 0.5 Hz
- 1.0 Hz
- π2HZ
- πHZ

**Q.**A man M slides down a curved frictionless track, starting from rest. The curve obeys the equation y=x22. The tangential acceleration of man is:

- g
- gx√x2+4
- g2
- gx√x2+1

**Q.**A particle is subjected to three SHMs in the same direction simultaneously each having equal amplitude a and equal time period. The phase of the second motion is 30 ahead of the first and the phase of the third motion is 30 ahead of the second. Find the amplitude of the resultant motion.

**Q.**A particle is subjected to two simple harmonic motions of the same frequency and direction. The amplitude of the first motion is 4.0 cm and that of the second is 3.0 cm. Find resultant amplitude when the phage difference is 1800. Find the resultant amplitude if the phase difference between the two motions is 0∘.

- 5cm
- 7cm
- 9cm
- 4cm

**Q.**Two simple harmonic motion of same amplitude, same frequency and a phase difference of π are superimposed at right angles to each other on a particle, the particle will describe a

- circle
- straight line
- ellipse
- figure of eight

**Q.**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

- vmax
- between 0 and vmax
- 0
- between 0 and −vmax

**Q.**From the differential equation, d2xdt2+16x=0

(1) we can find time period of oscillation

(2) we cannot find time period of oscillation

(3) we cannot find initial phase of oscillation

(4) we cannot find frequency of oscillation.

- 1 & 2
- 2 & 3
- 1 & 3
- 1, 2 and 3

**Q.**Two particles execute SHM with amplitude A and 2A and angular frequency ω and 2ω respectively. At t=0 they starts with some initial phase difference. At t=2π3ω they are in same phase. Their initial phase difference is

- 2π3
- π
- π3
- 4π3

**Q.**

If an SHM is represented by d2Xdt2+ax=0, its time period is

2πα

2πα

2π√a

2π√α