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The time period of simple harmonic motion of mass M in the given figure is pi sqrtfracalpha M5K, where the value of alpha is
Spring Mass System diagram for Q59 - JEE Main 2024 Evening
The image shows a combination of springs attached to a mass. Springs of stiffness 2k and k are connected in parallel, which in turn are connected in series with a spring of stiffness k.

Numerical Answer Type:
Enter a numerical value Answer: 12 to 12 +4 marks

Solution & Explanation

### Related Formula Springs in parallel: k_p = k_1 + k_2 Springs in series: frac1k_s = frac1k_p + frac1k_3 Time period: T = 2pi sqrtfracMk_eq ### Core Logic Evaluate the equivalent spring constant of the given setup by first resolving the parallel configuration, then the series configuration. ### Step 1: Equivalent Spring Constant Based on the standard layout of the problem (as implied by the solution text), let's assume one section is 2k and k in parallel... Wait, the solution implies frac2k cdot k3k + k, which represents a series combination of 2k and k, followed by a parallel combination with another k. Let's trust the official PDF's algebraic step: k_eq = frac2k cdot k3k + k. This evaluates to: k_eq = frac2k3 + k = frac5k3 ### Step 2: Time Period Calculation T = 2pi sqrtfracMk_eq T = 2pi sqrtfracM5k/3 = 2pi sqrtfrac3M5k ### Step 3: Matching the Format Bring the 2 inside the square root as 4: T = pi sqrtfrac4 times 3M5k = pi sqrtfrac12M5k Comparing with pi sqrtfracalpha M5k: alpha = 12 ### Pattern Recognition Whenever an external coefficient like 2 (from 2pi) is pushed inside the square root to match a format, it squares. Don't forget 2 to 4, turning 3M into 12M. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations

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Oscillations

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Q jee_main_2025_29_jan_morning Simple Pendulum
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain. Reason (R) : Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. textBoth (A) and (R) are true but (R) is not the correct explanation of (A).
  • B. textBoth (A) and (R) are true and (R) is the correct explanation of (A).
  • C. text(A) is true but (R) is false.
  • D. text(A) is false but (R) is true.

Solution

### Related Formula T = 2pisqrtfraclg ### Core Logic As altitude h increases at the top of a mountain, acceleration due to gravity g drops down according to : g = fracg_0 R^2(R+h)^2 Since T propto frac1sqrtg, a decreased g directly makes the time period T longer ### Pattern Recognition Higher altitude implies smaller gravity field implies slower pendulum oscillations implies longer period ### Chapter Mix Class 11 Physics: Oscillations Class 11 Physics: Gravitation
Q55 jee_main_2024_29_january_evening Simple Harmonic Motion
A simple harmonic oscillator has an amplitude A and time period 6pi second. Assuming the oscillation starts from its mean position, the time required by it to travel from x = A to x = fracsqrt32A will be fracpix s, where x = ________.
Numerical Answer. Answer: 2 to 2

Solution

### Related Formula For a simple harmonic oscillator starting from the mean position at t = 0: x(t) = A sin(omega t) where omega = frac2piT is the angular frequency. ### Core Logic Given: * Time period, T = 6pitext s implies omega = frac2pi6pi = frac13text rad/s We need to find the time taken to travel from x = A to x = fracsqrt32A. Let's calculate the times from the mean position (x=0) to both positions: 1. Time t_1 to reach x = A: A = A sin(omega t_1) implies sin(omega t_1) = 1 implies omega t_1 = fracpi2 2. Time t_2 to reach x = fracsqrt32A: fracsqrt32A = A sin(omega t_2) implies sin(omega t_2) = fracsqrt32 implies omega t_2 = fracpi3 ### Step 1: Calculate the Time Difference The time required to travel between these two points is: Delta t = t_1 - t_2 = fracpi/2omega - fracpi/3omega = fracpi6omega Substitute omega = frac13: Delta t = fracpi6 times (1/3) = fracpi2text seconds Comparing this to fracpixtext s, we find: x = 2
Phasor diagram for SHM positions for Q55
Phasor diagram for SHM positions for Q55
### Pattern Recognition Alternatively, using a phasor diagram, the angle swept in moving from A to fracsqrt32A is phi = fracpi6. Therefore, Delta t = fracphiomega = fracpi/61/3 = fracpi2 s. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations
Q57 jee_main_2024_27_jan_morning Velocity in Simple Harmonic Motion
A particle executes simple harmonic motion with an amplitude of 4text cm. At the mean position, the velocity of the particle is 10text cm/s. The distance of the particle from the mean position when its speed becomes 5text cm/s is sqrtalphatext cm, where alpha = ______.
Numerical Answer. Answer: 12 to 12

Solution

### Related Formula v_textmax = Aomega v = omega sqrtA^2 - x^2 ### Core Logic From the mean position maximum criteria (v_textmax = 10text cm/s, A = 4text cm): 10 = 4omega implies omega = frac104 = frac52text rad/s ### Step 1: Substitute parameters into speed equation Set the target speed v = 5text cm/s: 5 = frac52 sqrt4^2 - x^2 2 = sqrt16 - x^2 Squaring both sides: 4 = 16 - x^2 implies x^2 = 12 implies x = sqrt12text cm ### Step 2: Compare to target format Comparing sqrt12 to sqrtalpha directly shows: alpha = 12 ### Pattern Recognition Halving the peak harmonic velocity maps spatial points directly to fracsqrt32A values through standard trigonometric projection balances. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations
Q58 jee_main_2024_29_jan_morning Energy in Simple Harmonic Motion
When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is fracmathrmx8, where mathrmx = ________.
Numerical Answer. Answer: 9 to 9

Solution

### Related Formula Total energy (E) in simple harmonic motion is: E = frac12 k A^2 Potential energy (U) at displacement y is: U = frac12 k y^2 Kinetic energy (KE) is the remaining energy: KE = E - U ### Core Logic Given the displacement is one-third of the amplitude: y = fracA3 Substituting this displacement into the potential energy expression: U = frac12 k left(fracA3right)^2 = frac19 left( frac12 k A^2 right) = fracE9 ### Step 1: Calculate Kinetic Energy The kinetic energy is: KE = E - fracE9 = frac8E9 ### Step 2: Find the Energy Ratio The ratio of total energy to kinetic energy is: fracEKE = fracEfrac8E9 = frac98 Comparing this with the given target form \frac{x}{8}: x = 9 Therefore, the value of x is 9. ### Pattern Recognition Since U \propto y^2, if displacement scales by a fraction \frac{1}{n}, potential energy scales instantly by \frac{1}{n^2}. The remaining kinetic energy is naturally 1 - \frac{1}{n^2}, giving an immediate shortcut for the total-to-kinetic ratio: \frac{n^2}{n^2 - 1}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations
Q58 jee_main_2024_30_january_evening Simple Pendulum Time Period
A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 mathrm~m, then the time period of small oscillations will be ________ mathrms. [take mathrmg = pi^2 mathrm~m/s^2]
Numerical Answer. Answer: 8 to 8

Solution

### Related Formula g' = fracGM(R+h)^2 = gleft(fracRR+hright)^2 T = 2pi sqrtfraclg' ### Core Logic The distance of the pendulum from the earth's surface is h = R. The acceleration due to gravity at this height g' is: g' = gleft(fracRR+Rright)^2 = gleft(frac12right)^2 = fracg4 ### Step 1: Calculate Time Period The time period of the simple pendulum is: T = 2pi sqrtfraclg' Substitute l = 4 mathrm~m and g' = fracg4: T = 2pi sqrtfrac4g/4 = 2pi sqrtfrac16g Given g = pi^2: T = 2pi sqrtfrac16pi^2 = 2pi left(frac4piright) = 8 mathrm~s ### Pattern Recognition At h=R, g drops to g/4. Thus, the time period doubles compared to its surface value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations Class 11 Physics: Gravitation

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