SHM – Time Period of Spring | Class 11 Physics + JEE Main

SHM – Time Period of Spring | Class 11 + JEE Main

Simple Harmonic Motion (SHM) Class 11 Physics ka ek important topic hai, especially JEE Main ke liye. Spring-block system mein time period calculate karna, mass aur spring constant ka effect samajhna, aur springs ki series-parallel combination solve karna is topic ke important concepts hain.

Is article mein hum spring ke time period ka formula step-by-step derive karenge, important relationships samjhenge, vertical spring ke concepts revise karenge aur JEE Main-style numerical questions solve karenge.

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1. SHM Kya Hota Hai?

Simple Harmonic Motion (SHM) ek periodic motion hai jisme restoring force equilibrium position se displacement ke directly proportional aur displacement ke opposite direction mein hota hai.

F = −kx

Where:

• F: Restoring force, measured in newtons (N).
• k: Spring constant, measured in N/m.
• x: Displacement from the equilibrium position, measured in metres (m).

Formula mein negative sign batata hai ki restoring force displacement ke opposite direction mein act karta hai. Force block ko equilibrium position ki taraf wapas laane ki koshish karta hai.

Example: Spring-Block System

Imagine karo ek block horizontal spring se attached hai. Agar block ko equilibrium position se pull karke release kiya jaye, toh spring block par restoring force lagati hai. Block aage-peeche oscillate karta hai.

Ideal spring ke liye Hooke's law follow hota hai:

F = −kx

Jab displacement zyada hota hai, toh restoring force ka magnitude bhi zyada hota hai, provided spring Hooke's law ke valid range mein kaam kar rahi ho.

JEE Main Tip: SHM identify karne ke liye check karo ki acceleration displacement ke proportional aur opposite direction mein hai ya nahi.

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2. Spring Ka Time Period: Formula Derivation

Spring-block system ka time period derive karne ke liye Newton's Second Law use karte hain.

Step 1: Newton's Second Law Apply Karo

Newton's Second Law ke according:

F = ma

Spring ka restoring force hai:

F = −kx

Dono equations ko combine karne par:

−kx = ma

Acceleration ko isolate karte hain:

a = −(k/m)x

Step 2: SHM Ke Standard Equation Se Compare Karo

Simple Harmonic Motion mein acceleration ka standard equation hota hai:

a = −ω²x

Yahan ω angular frequency hai, jiska unit rad/s hota hai.

Dono acceleration equations compare karne par:

−(k/m)x = −ω²x

Therefore:

ω² = k/m

ω = √(k/m)

Step 3: Time Period Calculate Karo

Angular frequency aur time period ke beech relationship hai:

ω = 2π/T

Isliye:

T = 2π/ω

Ab ω = √(k/m) substitute karte hain:

T = 2π√(m/k)

Ye ideal spring-block system ke time period ka most important formula hai.

Formula Mein Variables

• T: Time period, seconds (s).
• m: Oscillating block ka mass, kilograms (kg).
• k: Spring constant, N/m.

Time period ka matlab hai ek complete oscillation ko finish karne mein lagne wala time.

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3. Mass Aur Spring Constant Ka Effect

Spring time period ka formula hai:

T = 2π√(m/k)

Is formula se clear hai ki time period mass ke square root ke directly proportional aur spring constant ke square root ke inversely proportional hota hai.

T ∝ √m

T ∝ 1/√k

Mass Increase Hone Par

Agar spring constant same rahe aur block ka mass increase ho, toh time period increase hota hai.

• Mass 4 times karne par time period 2 times ho jayega.
• Mass 9 times karne par time period 3 times ho jayega.
• Mass 16 times karne par time period 4 times ho jayega.

Example 1: Mass Four Times

Ek spring-block system ka initial time period T hai. Agar mass ko 4m kar diya jaye aur spring constant same rahe, toh new time period:

T' = 2π√(4m/k)

T' = 2 × 2π√(m/k)

T' = 2T

Spring Constant Increase Hone Par

Agar mass same rahe aur spring constant increase ho, toh time period decrease hota hai.

• Spring constant 4 times karne par time period half ho jayega.
• Spring constant 9 times karne par time period one-third ho jayega.
• Spring constant 16 times karne par time period one-fourth ho jayega.

Example 2: Spring Constant Four Times

Agar spring constant k se 4k ho jaye, toh:

T' = 2π√(m/4k)

T' = (1/2) × 2π√(m/k)

T' = T/2

JEE Main Shortcut: Mass ka factor square root ke through aata hai, aur spring constant ka factor denominator mein square root ke through aata hai.

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4. Springs in Series

Jab do springs ko end-to-end connect kiya jata hai, toh unhe series combination kehte hain. Is combination ka effective spring constant individual spring constants se different hota hai.

Agar individual spring constants k₁ aur k₂ hain, toh series combination ke liye:

1/k_eq = 1/k₁ + 1/k₂

Is equation ko simplify karne par:

k_eq = (k₁k₂)/(k₁ + k₂)

Equal Springs in Series

Agar dono springs ka spring constant equal ho:

k₁ = k₂ = k

Then:

1/k_eq = 1/k + 1/k = 2/k

k_eq = k/2

Ab effective spring constant ko time period ke formula mein substitute karte hain:

T = 2π√(m/k_eq)

T = 2π√(m/(k/2))

T = 2π√(2m/k)

Equal springs ki series combination ka effective spring constant k/2 hota hai, isliye single spring ke comparison mein time period increase hota hai by a factor of √2.

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5. Springs in Parallel

Jab springs ko parallel arrangement mein connect kiya jata hai aur dono same block ko support karte hain, toh ideal setup mein unka effective spring constant add hota hai.

k_eq = k₁ + k₂

Equal Springs in Parallel

Agar dono springs ka spring constant k hai:

k_eq = k + k

k_eq = 2k

Time period:

T = 2π√(m/k_eq)

T = 2π√(m/2k)

T = 2π√(m/2k)

Equal springs ki parallel combination ka effective spring constant 2k hota hai. Isliye time period single spring ke comparison mein 1/√2 times ho jata hai.

Series vs Parallel: Quick Comparison

Combination	Effective Spring Constant	Equal Springs
Series	1/keq = 1/k1 + 1/k2	keq = k/2
Parallel	keq = k1 + k2	keq = 2k

Shortcut: Series combination mein effective spring constant decrease hota hai, jabki parallel combination mein increase hota hai.

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6. Vertical Spring and Time Period

Jab ek mass ko vertical spring se hang kiya jata hai, toh gravity ke karan spring stretch hoti hai. Is stretching ki wajah se equilibrium position shift ho jati hai.

Step 1: Equilibrium Position Find Karo

Equilibrium position par spring force aur gravitational force balance karte hain.

kx₀ = mg

Isliye equilibrium extension:

x₀ = mg/k

Yahan x₀ natural length se static extension ko represent karta hai.

Step 2: Equilibrium Ke Around Oscillation

Mass ko equilibrium position se thoda displace karke release karne par woh equilibrium ke around oscillate karta hai. Ideal spring ke liye oscillation ka time period hota hai:

T = 2π√(m/k)

Is formula mein gravitational acceleration g directly appear nahi karta.

Does Gravity Affect the Time Period?

Ideal spring-block system mein gravity equilibrium position change karti hai, lekin equilibrium ke around small oscillations ka time period change nahi karti, provided spring constant aur mass same rahen.

Important distinction: Gravity static extension ko affect karti hai, jabki ideal time period mass aur spring constant par depend karta hai.

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7. Solved JEE Main Question

Question:

Ek 2 kg block ko 200 N/m spring se attach kiya gaya hai. Ideal spring-block system ka time period find karo.

Given

• Mass, m = 2 kg
• Spring constant, k = 200 N/m

Step 1: Formula Write Karo

T = 2π√(m/k)

Step 2: Values Substitute Karo

T = 2π√(2/200)

T = 2π√(0.01)

T = 2π × 0.1

Final Answer

T = 0.2π s

T ≈ 0.628 s

Isliye block ko ek complete oscillation karne mein approximately 0.628 seconds lagenge.

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8. Additional JEE Main Practice Examples

Example 1: Mass Increase

Ek spring-block system ka time period 2 seconds hai. Agar mass 4 times kar diya jaye aur spring constant same rahe, toh new time period find karo.

Solution:

T' = T√(m'/m)

T' = 2√4 = 4 s

Answer: 4 seconds.

Example 2: Spring Constant Increase

Ek spring-block system ka time period 6 seconds hai. Agar spring constant 9 times kar diya jaye aur mass same rahe, toh new time period find karo.

Solution:

T' = T√(k/k')

T' = 6√(1/9)

T' = 6/3 = 2 s

Answer: 2 seconds.

Example 3: Equal Springs in Series

Ek block mass m ki ek spring constant k wali spring se attached hai. Agar same spring constant ki do springs series mein connect kar di jayein, toh time period single spring ke comparison mein kitna hoga?

Solution:

k_eq = k/2

T' = 2π√(m/(k/2))

T' = √2 × 2π√(m/k)

T' = √2 T

Answer: Time period √2 times ho jayega.

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9. Common Mistakes to Avoid

1. Formula ulta likhna: Correct formula T = 2π√(m/k) hai, not 2π√(k/m).
2. Mass ka factor galat apply karna: Mass 4 times hone par time period 4 times nahi, 2 times hota hai.
3. Spring constant ka effect bhoolna: Spring constant 4 times hone par time period half ho jata hai.
4. Series aur parallel formulas interchange karna: Series mein reciprocal relation aur parallel mein direct addition use hota hai.
5. Vertical spring mein gravity ko directly time period mein include karna: Ideal spring ke liye T = 2π√(m/k), aur g directly formula mein nahi aata.
6. Units ignore karna: Mass ko kg aur spring constant ko N/m mein use karo.
7. Angular frequency aur frequency confuse karna: Angular frequency ω = √(k/m), jabki frequency f = ω/(2π).

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10. Important Formula Sheet

Concept	Formula
Restoring Force	F = −kx
Acceleration	a = −(k/m)x
Angular Frequency	ω = √(k/m)
Time Period	T = 2π√(m/k)
Frequency	f = (1/2π)√(k/m)
Mass Dependence	T ∝ √m
Spring Constant Dependence	T ∝ 1/√k
Series Springs	1/keq = 1/k1 + 1/k2
Parallel Springs	keq = k1 + k2
Vertical Spring Equilibrium	x0 = mg/k
Vertical Spring Time Period	T = 2π√(m/k)

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11. One-Minute Revision

• SHM mein restoring force displacement ke proportional aur opposite direction mein hota hai.
• Restoring force: F = −kx.
• Angular frequency: ω = √(k/m).
• Spring time period: T = 2π√(m/k).
• Frequency: f = (1/2π)√(k/m).
• Mass 4 times → time period 2 times.
• Mass 9 times → time period 3 times.
• Spring constant 4 times → time period half.
• Series springs: 1/k_eq = 1/k₁ + 1/k₂.
• Parallel springs: k_eq = k₁ + k₂.
• Equal springs in series: k_eq = k/2.
• Equal springs in parallel: k_eq = 2k.
• Vertical spring equilibrium extension: x₀ = mg/k.
• Ideal vertical spring ka time period gravity se directly depend nahi karta.

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12. Practice Questions for JEE Main

Question 1

A block of mass 1 kg is attached to a spring of spring constant 100 N/m. Find its time period.

A) 0.1π s
B) 0.2π s
C) 0.5π s
D) 2π s

Solution:

T = 2π√(1/100) = 2π × 0.1

Answer: B) 0.2π s

Question 2

If the mass attached to a spring is increased to 9 times its original value, the new time period becomes:

A) T/3
B) 3T
C) 9T
D) T

Answer: B) 3T

Question 3

Two identical springs, each having spring constant k, are connected in series. What is the effective spring constant?

A) 2k
B) k
C) k/2
D) k/4

Answer: C) k/2

Question 4

Two identical springs, each having spring constant k, are connected in parallel. What is the effective spring constant?

A) k/2
B) k
C) 2k
D) 4k

Answer: C) 2k

Question 5

A mass is suspended vertically from an ideal spring and oscillates about its equilibrium position. Which quantity directly determines the time period in the ideal model?

A) Only gravitational acceleration
B) Mass and spring constant
C) Only the equilibrium extension
D) Only gravitational force

Answer: B) Mass and spring constant

Question 6

If the spring constant becomes 16 times its original value while mass remains unchanged, the new time period is:

A) 4T
B) 2T
C) T/2
D) T/4

Answer: D) T/4

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13. Download / View SHM Notes PDF

SHM aur spring time period ke notes ko revise karne ke liye neeche apni PDF file ka link add karein. Agar PDF Blogger ya kisi publicly accessible hosting platform par uploaded hai, toh uska direct URL iframe mein use kar sakte hain.

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14. Frequently Asked Questions (FAQs)

Q1. What is the time period of a spring-block system?

An ideal spring-block system has time period T = 2π√(m/k), where m is the oscillating mass and k is the spring constant.

Q2. What happens to the time period when mass increases?

Time period mass ke square root ke proportional hota hai. Agar mass 4 times ho jaye, toh time period 2 times ho jata hai.

Q3. What happens when spring constant increases?

Time period spring constant ke square root ke inversely proportional hota hai. Agar spring constant 4 times ho jaye, toh time period half ho jata hai.

Q4. What is the effective spring constant for springs in series?

Two springs in series ke liye 1/k_eq = 1/k₁ + 1/k₂.

Q5. What is the effective spring constant for springs in parallel?

Two springs in parallel ke liye k_eq = k₁ + k₂.

Q6. Does gravity affect the time period of a vertical spring?

Ideal spring-block system mein gravity equilibrium position ko shift karti hai, lekin equilibrium ke around small oscillations ka time period T = 2π√(m/k) hi rehta hai.

Q7. Is SHM included in JEE Main Physics?

SHM and spring-block oscillations are important concepts for JEE Main Physics preparation. Apne target examination year ke liye latest official NTA syllabus check karein to confirm the exact scope of Oscillations and Waves.

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Final Thoughts

Spring-block SHM ko master karne ke liye sabse pehle restoring force ka concept samjho, phir Newton's Second Law se angular frequency aur time period derive karo. Uske baad mass aur spring constant ke changes, series-parallel combinations aur vertical spring ke questions practise karo.

JEE Main ke liye formula ko blindly memorise karne ke bajaye uski dependence samajhna zyada useful hai. Yaad rakho: mass badhne se time period badhta hai, spring constant badhne se time period kam hota hai, aur ideal vertical spring ke time period mein gravity directly appear nahi karti.

Regular numerical practice se aap spring time period ke direct aur conceptual questions ko accurately solve kar sakte ho.